3-Dimensional Rotation Space
Date: 05/18/2009 at 19:06:10 From: Seb Subject: 3-dimensional rotation space One can easily embed the space of 3-dimensional rotations in 3d Euclidean space (just represent a rotation by a vector whose magnitude gives the angle of rotation, and whose direction gives the axis). Doing this leads one to a sphere of radius \pi whose antipodal points are identified (This is because rotations of \pi are equivalent to flipping the rotation axis). I think this is RP^3. Now the question is to consider a closed loop representing a rotation of 2\pi, and prove that one cannot continuously deform this loop to a point. I am a physicist, and none of the math I have learned seems useful for this type of question!! I have a suspicion that homotopy theory would be useful, but know very little of it, and the Algebraic Topology textbooks I have looked at are scary! Are there any "elementary" methods one can use?
Date: 05/20/2009 at 06:56:14 From: Doctor Jacques Subject: Re: 3-dimensional rotation space Hi Seb, This is a question about homotopy, and any rigorous proof should involve some algebraic topology (I agree that "scary" is an appropriate word). It is still possible to give a (non rigorous) "plausibility argument". Let us first consider the two-dimensional case (by this, I mean the projective plane RP^2, NOT the group of two-dimensional rotations). From a topological point of view, this surface is homeomorphic to a disc with the antipodal points of the boundary identified (in what follows, I will draw the disc as a square, which is much easier using ASCII characters; in the "rubber world" of topology, this is obviously equivalent). You are probably familiar with the use of this plane in coordinate geometry (homogeneous coordinates); the boundary of the disc corresponds to the line at infinity. We want to show that the fundamental group of RP^2 is not trivial, i.e., there is a closed path that cannot be continuously deformed to a point. Consider the path around O in the following diagram: +-----------+ | | | O | A' +--->-+-->--+ A | 2 1 | | | +-----------+ Figure 1 (In this diagram, the points A and A' are actually the same point of RP^2). I claim that it is not possible to deform this path continuously to a point, while keeping the point O fixed. To see why, consider what happens when a path that is fully enclosed in the square is deformed and crosses the boundary: +-----------+ B' + -----+ A |\ / | | | O | |/ \ | A' + -----+ B +-----------+ Figure 2 When the path reaches the boundary of the square, it becomes tangent to it; just after that, you have the configuration described in figure 2. Note the identifications A = A' and B = B'. Because the points A and B are produced continuously from a single point (the point of tangency), they can be arbitrarily close to each other, and cannot therefore be antipodal. This means that the points A = A' and B = B' are distinct points of the projective plane. The conclusion is that points on the boundary can only be created or destroyed in pairs. As you have only one point A = A' on the boundary in the original path (figure 1), you will always have an odd number of points (except when the path is tangent to the boundary, but a point of tangency is actually two points). It is therefore not possible to reduce the path to a single point, because you would then have no points on the boundary. There is also another representation of RP^2 that sheds additional light on what happens. We can consider RP^2 as a sphere S^2 (the surface) with antipodal points identified. The disc used in the previous representation corresponds to a hemisphere. Specifically, a point (x,y) inside the unit disc corresponds to the pair of antipodal points (x, y, sqrt(1 - x^2 - y^2)) (-x, -y, -sqrt(1 - x^2 - y^2)) on the unit sphere. In that representation, the path of figure 1 corresponds to an open path from the North pole to the South pole on the sphere. As the endpoints of that path are distinct on the sphere (before antipodal identification), you cannot deform the path continuously to a small path around a single point (you must always cross the equator an odd number of times). The crucial point here (where a rigorous demonstration would be required) is that a continuous deformation of a path in RP^2 corresponds to a single continuous deformation of the corresponding path on the sphere. You can use exactly the same argument for the projective space RP^3, which, as you correctly state, is homeomorphic to the group of rotations of Euclidean 3-space, SO(3). RP^3 can be seen as the unit ball in R^3 (or a ball with radius pi, but this is only a matter of units), with antipodal points identified. A rotation of 2*pi corresponds to a closed path that crosses the boundary of the ball once. The same argument shows that, when you deform that path continuously, you will always cross the boundary an odd number of times (in particular, at least once). For the second interpretation, you must consider the unit ball of R^3 as half the (hyper-)surface of S^3, the unit sphere in four dimensions. The point (x,y,z) of the unit ball corresponds to the pair of antipodal points: (x, y, z, t) (-x, -y, -z, -t) on S^3, with t = sqrt(1 - x^2 - y^2 -z^2). As the argument above is based on the fact that you create points on the boundary in pairs, this raises naturally the question of a rotation of 4*pi (720°). It is interesting that such a rotation can indeed be continuously deformed to a point. To see this, we start again with the simpler case of RP^2. We consider a closed path that crosses the boundary twice: B' +--------+--------+ | | | | 4v | | | O | A' +-->-----+--->----+ A | 2 | 1 | | 3v | | | | +--------+--------+ B Figure 3 We start from O, follow the path (1 2) to the left through A = A', then the path (3 4) downwards through B = B'. Note that O represents 4 points: the starting point O1, the intermediate points O2 (the end of the first path) and O3 (the start of the second path) and the final point O4. The combined path is from O1 to O4; in a deformation, these points must be left fixed. On the other hand, the points O2 and O3 are now intermediate points on the path, and we are free to move them (together, because of continuity). We can deform the path in this way: +-----------------+ | 4 | A' +->2 --<----+ B | \ / | | | O | | / \ | B' +-<3 -->----+ A | 1 | +-----------------+ Figure 4 and it is easy to see that you can now push the leftmost part (2 3) through the boundary, close the path on the right, and deform it to a point. As before, the same argument applies to RP^3 (it is a property of projective spaces, not of rotation groups). However, as RP^3 is homeomorphic to SO(3) this has surprising consequences, one of which is illustrated by R. Feynman's "plate trick" : it is possible to hold a plate on your hand, and, by a rather complicated movement, make it do two complete revolutions (a rotation of 720°) while still keeping its face up and without dislocating your shoulder joint (at least, not too much :)). See, for example: Wikipedia: Plate trick http://en.wikipedia.org/wiki/Plate_trick Math Pages: Dirac's Belt http://www.mathpages.com/home/kmath619/kmath619.htm In topological terms, the fundamental group of RP^n is Z_2 (the integers modulo 2) for all n >= 2. The argument breaks down for n = 1 (the projective line RP^1 is the same as the circle S^1), because, roughly speaking, there is no room to move the points O2 = O3. In physics, this is deeply related to the behavior of spin 1/2 particles (as you are a physicist, this is maybe the reason why you ask this). Please feel free to write back (using the URL in this message) if you want to discuss this further. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/
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