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Continuous R^2xR^2xR^2/E^+(2) -> R^3 injection

  1. Jan 1, 2013 #1
    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 rotation-invariant 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 rigid-body 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 translation-independent 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. jcsd
  3. Jan 6, 2013 #2
    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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