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This is a question that comes from my (biological) research. I'm very weak in 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)). You can use the translation degree of freedom to reduce this to ℝ^2×ℝ^2/SO(2).

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 suspect, is impossible. Am I right?

Thanks for any help.

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  • $\begingroup$ Maybe I'm misunderstanding, but it sounds like side-angle-side from Euclidean geometry should do the trick. The first side being the distance from point 1 to point 2, the second side being the distance from point 2 to point 3, and the angle being the oriented angle which swings the ray 1 (originating at point 2 and passing through point 1) to ray 2 (originating at point 2 and passing through point 3), swinging in the counterclockwise direction. The data (S, A, S) gives you the point in $\mathbb{R}^3$. Does this seem right? $\endgroup$
    – Todd Trimble
    Commented Jan 5, 2013 at 18:55
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    $\begingroup$ (Or perhaps even more simply, side-side-side.) $\endgroup$
    – Todd Trimble
    Commented Jan 5, 2013 at 19:01
  • $\begingroup$ unknown google -- if I understand you correctly, your quotient space is juet the space of all triangles in $\mathbb{R}^2$ with ordered vertices; it includes degenerate triangles, in which one of the vertices lies in the interior of the segment that joins the other two. This space is indeed a subspace of $\mathbb{R}^3$: each triangle is determined, up to a composition of rotations, reflections and translations, by the lengths of the edges (these are ordered, as the vertices are). If you do not allow reflections, then a triangle is determined by the lenghts of the sides plus orientation. $\endgroup$
    – algori
    Commented Jan 5, 2013 at 19:08
  • $\begingroup$ @algori: I could have sworn I just made a similar comment... $\endgroup$
    – Todd Trimble
    Commented Jan 5, 2013 at 19:20
  • $\begingroup$ Todd -- so you did but I didn't see it when I started writing mine. $\endgroup$
    – algori
    Commented Jan 5, 2013 at 19:29

2 Answers 2

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By translation, fix the first point to be at the origin. Consider the other two points as complex numbers, and take their quotient. As long as the other two points are not both at the origin, this continuously gives an element of $\mathbb{CP}^1\cong S^2$. Up to rotation, the other two points are determined by this quotient together with a scale parameter, such as the sum the sides of the triangle. A point on the sphere together with a nonzero scale parameter gives you a point in $\mathbb{R}^3\setminus\{0\}$. Thus we have a continuous injection (in fact, homeomorphism) from your space (except for the point where all three points coincide) to $\mathbb{R}^3\setminus\{0\}$.

What about the case when all three points coincide? Well, a sequence in your space will converge to the case when all three points coincide iff the scale parameter converges to 0. Thus we can continuously extend the map to send that point to $0\in\mathbb{R}^3$. We thus get that your space is actually homeomorphic to $\mathbb{R}^3$.

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  • $\begingroup$ This seems to work. Good! $\endgroup$
    – Todd Trimble
    Commented Jan 5, 2013 at 20:13
  • $\begingroup$ OK, right. Here's what I have. Not exactly what you propose for some practical reasons having to do with the application, but the same idea: $v_1 = z_2 - z_1$, $v_2 = z_3 - z_2$, $(\rho,\phi) = (|v_1/v_2|, arg(v_1/v_2))$, $r = |v_1| + |v_2|$, $\theta = \pi - 2 tan^-1(\rho)$, Convert $(r,\phi,\theta)$ to (x,y,z) in the usual way. Great! I'll give this a try. Muchas gracias. Umm... dumb question: How the Hell do I create a line break in a comment? $\endgroup$
    – Leon Avery
    Commented Jan 5, 2013 at 21:01
  • $\begingroup$ Hmm. I already see the practical problem with this: the information about speed of movement ends up spread out over x, y, and z, and I really want it in a single number. Ah, well. That's biology, which is my problem. You answered my question. $\endgroup$
    – Leon Avery
    Commented Jan 5, 2013 at 21:14
  • $\begingroup$ So, following algori's last comment above, if I decide to yank out the speed, which is the scale factor, I'm left with $S^2$, which by Borsuk-Ulam explains why I was always left with 1 bit of missing info when I tried to map the remaining two degrees of freedom to $\mathbb{R}^2$. $\endgroup$
    – Leon Avery
    Commented Jan 5, 2013 at 21:33
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Let $v_1=z_1-z_3,v_2=z_2-z_3$ and $p(v_1,v_2)=(2v_1\bar v_2, |v_1|^2-|v_2|^2)\in\mathbb{C}\times\mathbb{R}.$ The map $(z_1,z_2,z_3)\mapsto p(v_1,v_2)$ defines a homeomorphism $\mathbb{C}^3/E^+(2)\cong\mathbb{R}^3$.

The map $p:\mathbb{C}^2\to\mathbb{R}^3$ has some nice properties which might be useful for your applications. You can see it in the Wikipedia article on the Hopf fibration (where the formula for $p$ is taken from).

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  • $\begingroup$ In fact, this turns out to be identical (except for sign changes) to the map I worked out as a result of Eric Wofsey's answer. Let's call $s_i = |v_i|$ and $\theta = arg(v_2/v_1)$. Then it turns out to be $(x,y,z) = (s_1 s_2 cos\theta, s_1 s_2 sin\theta, s_2^2-s_1^2)$. This makes more physical sense if I use square roots: $(\sqrt{s_1 s_2} cos\theta, \sqrt{s_1 s_2} sin\theta, s_2-s_1)$, which, if z is a smooth function of time and the sampling interval is small, are roughly mean speed, radial acceleration, and tangential acceleration. $\endgroup$
    – Leon Avery
    Commented Jan 6, 2013 at 12:40
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    $\begingroup$ Actually, there is more then a sign change. The map I wrote is, in your notation, $(x,y,z) = (2s_1 s_2 cos\theta, 2s_1 s_2 sin\theta, s_1^2-s_2^2)$, i.e. there is an extra factor of 2 in the $x$ and $y$ coordinates, but not in the $z$ coordinate. This has the following effect: $p:\mathbb{C}^2\to\mathbb{R}^3$ maps the sphere $|v_1|^2+|v_2|^2=r^2$ onto the sphere $x^2+y^2+z^2=r^4.$ $\endgroup$
    – Gil Bor
    Commented Jan 7, 2013 at 0:39

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