0
$\begingroup$

I have 2 points $A=(x_A,y_A), B=(x_B,y_B)$ on a unit circle $O$. The distance between $A$ and $B$ goes through the perimeter of the circle. How can I transform this space to a space with higher dimensions where the distances can be computed using Euclidean formula, and the original distances are preserved as much as possible? In fact, I don't know what is the main field of math concerning such transformations.

Your help is appreciated.

$\endgroup$
  • $\begingroup$ You want to transform a circle in the plane to a circle in a higher dimensional space? Or you are trying to transform it into a higher dimensional object such as a sphere? $\endgroup$ – Morgan Rodgers Nov 4 '15 at 7:14
  • $\begingroup$ None of the above. I need a higher dimension space where euclidean distances are equal to the distances in the original space, i.e., the arc length. $\endgroup$ – remo Nov 4 '15 at 19:34
  • $\begingroup$ It's impossible to match distance exactly if the image curve (in the high-dimensional space) is smooth, since the "intrinsic" distance on a circle isn't smooth "at the antipodes". If the goal is a "best approximation", a precise quantification of the total error is essential: How exactly are you measuring "quality of fit"? (You might search for "isometric embeddings of metric spaces", as a branch of analysis.) There does exist an isometric embedding into the (infinite-dimensional) space of continuous functions on the circle; this is well-known, but I don't have a reference at hand. $\endgroup$ – Andrew D. Hwang Nov 5 '15 at 11:17
2
$\begingroup$

Hint:

Map the points using the transform

$$\tan(z)=\frac yx,$$ where $z$ is evaluated on four quadrants, and the distance between $A$ and $B$ turns to the Euclidean $$|z_A-z_B|.$$

Unfortunately, an essential nonlinearity remains because of phase wraparound, and the exact formula must be

$$\pi-||z_A-z_B|-\pi|.$$

$\endgroup$
  • $\begingroup$ This seems interesting. Would you please how can I extend this to a sphere, i.e., $A=(x_A,y_A,z_A)$. $\endgroup$ – remo Nov 5 '15 at 20:39
  • $\begingroup$ I don't think that there are distance-preserving projections of the sphere. See this: geokov.com/education/map-projection.aspx $\endgroup$ – Yves Daoust Nov 5 '15 at 20:44
  • $\begingroup$ These mappings try to map to a paper (2D), while i need a higher dimensional space but in Euclidean space. $\endgroup$ – remo Nov 5 '15 at 20:46
  • $\begingroup$ @remo The mapping I suggested for the circle doesn't work in a higher dimensional space, on the opposite. $\endgroup$ – Yves Daoust Nov 5 '15 at 20:49
2
$\begingroup$

You have to keep in mind that Euclidean distances do not behave like arc length distances, so it is unlikely that you will be able to do this.

Notice that you will need to map the points of the circle to some closed curve. The fact that you need the distances of opposite points of the circle preserved, means for every point $\mathbf{x}$ on your closed curve you need to have a unique point at distance $\pi$ from $\mathbf{x}$. So you will be mapping your circle to some closed curve on an $n$-sphere (this is just an $n$-dimensional analog of the sphere in $n$ dimensions) having diameter $\pi$.

I'll admit I'm not sure exactly where it will break down, or what the optimal way to map it will be.

$\endgroup$
  • $\begingroup$ Do you mean I cannot embed the distanes into an Eulidean space? If so, what is the best approximation? $\endgroup$ – remo Nov 5 '15 at 8:51
  • 1
    $\begingroup$ I don't know. Possibly there is a way to embed it that I am overlooking, but I would be surprised. I'm not sure what you are trying to do, or why you would try to estimate these with Euclidean distances. It is possible that your best approximation would be to transform it into a circle with radius $\frac{\pi}{2}$, at least then the Euclidean distance would match up for opposite points. $\endgroup$ – Morgan Rodgers Nov 5 '15 at 9:19
  • $\begingroup$ Please note I have a number of predefined points on the circle, not the circle itself. So I think there must be some optimum point. Have a look at the more formal version but in 3d. [math.stackexchange.com/questions/1529600/… $\endgroup$ – remo Nov 15 '15 at 20:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.