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Continuous R^2xR^2xR^2/E^+(2) > R^3 injection 
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#1
Jan113, 03:35 PM

P: 455

This is a question that comes from my research. I know next to nothing about topology, so I'm not able to assure myself of the answer. The problem is this: I'm watching an animal move in two dimensions. At three successive points in time I have three positions, (x1,y1), (x2,y2), (x3,y3). But there are three uninteresting degrees of freedom in these numbers: two that say where it all happened and one that gives the angle you're looking at it from. In other words, I am only interested in translation and rotationinvariant aspects of the motion. Thus, the three positions are best understood not as being a point in ℝ^2×ℝ^2×ℝ^2, but in the orbit space ℝ^2×ℝ^2×ℝ^2/E+(2), E+(2) being the group of rigidbody motions in two dimensions, acting uniformly on all three positions, i.e. e in E+(2) acts on ((x1,y1), (x2,y2), (x3,y3)) to produce (e(x1,y1), e(x2,y2), e(x3,y3)).
I want to get three numbers that contain all the rotation and translationindependent information in (x1,y1), (x2,y2), (x3,y3). This is easy. I would also like the mapping to be continuous. That is, I would like to have a continuous injection from ℝ^2×ℝ^2×ℝ^2/E+(2) > ℝ^3. This, I believe, is impossible. Am I right? I have a feeling this is basically Borsuk–Ulam, but like I said, I'm pretty ignorant of topology. Thanks for any help. 


#2
Jan613, 06:16 AM

P: 455

Got some answers to this on mathoverflow.net. It turns out there ARE continuous bijections ℝ^2×ℝ^2×ℝ^2/E+(2) > ℝ^3.



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