Isometry between euclidian and noneuclidian spaceby giokara Tags: euclidian, isometry, noneuclidian, space 

#1
Jul2411, 10:26 AM

P: 9

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 noneuclidian 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}{1D}} [/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 noneuclidian 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.. 


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