Isometry between euclidian and non-euclidian space

[giokara]

Hi

For the moment, I am working on dimensionality reduction of data and the following question has risen during my work.
I have a set of data in a non-euclidian space, lying on a manifold for which I know the metric measure d between points with coordinates (mu_1,sigma_1) and (mu_2,sigma_2):

[itex] D((\mu_1,\sigma_1);(\mu_2,\sigma_2)) = \sqrt {\frac{(\mu_1-\mu_2)^2+2(\sigma_1-\sigma_2)^2}{(\mu_1-\mu_2)^2+2(\sigma_1+\sigma_2)^2}}[/itex]
[itex]d((\mu_1,\sigma_1);(\mu_2,\sigma_2)) = 2\log^2{\frac{1+D}{1-D}} [/itex]

A problem arises when I try to find the dimension. The algorithm that I use finds a configuration of points in a lower dimensional Euclidian space which conserves the distances between the points. From this, I should be able to find the real dimensionality of the data. However, the algorithm seems not to perform very well (it still is more or less able to find the correct dimension with fictive data for which I know the true dimensionality, however, the results are almost inconclusive).
So I wondered, is it always possible to find an isometric transformation between a region in a non-euclidian space and a region in a Euclidian space? Perhaps it is simply not possible to use the current approach to my problem.

PS: Excuse me for the poor terminology I use. This stuff is all very, very new to me..

 

[pat_connell]

:)Thanks in advance for your input. Yes, it is always possible to find an isometric transformation between a region in a non-euclidean space and a region in a Euclidean space. This is known as the Riemannian Isometry Theorem. The theorem states that any two regions of the same size in a non-Euclidean space can be mapped to each other using an isometric transformation. In your case, you should be able to apply the Riemannian Isometry Theorem to map your data from its non-Euclidean space to a lower-dimensional Euclidean space while preserving the metric measure of distance. This should allow you to accurately determine the dimensionality of your data set.