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 Exercise [12.17] 
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Joined: 07 May 2009, 16:45
Posts: 62
Post Exercise [12.17]
By representing a rotation in ordinary 3-space as a vector pointing along the rotation axis of length equal to the angle of rotation, show that the topology of \boldsymbol R can be described as a solid ball (of radius \pi) bounded by an ordinary sphere, where each point of the sphere is identified with its antipodal point. Give a direct argument to show why a closed loop representing a 2\pi-rotation canot. continuously deformed to a point.
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Although this exercise is at the end of chapter 12 and so has a large number (12.17), it is referenced in an end note that refers back to section 12.1. So it really goes with the start of the chapter.

The \boldsymbol R mentioned in the exercise is the space representing the rotation of a body. I have tried to illustrate the representation of rotation mentioned in the exercise in the following diagram.

Image

The rotation is in the about the Z axis. I try to show how various rotations of the X axis in the X-Y plane is shown as vectors in the Z or -Z direction. For instance \alpha is a small rotation in the negative direction and is represented by small vector in the negative Z direction. Rotations \beta and \delta are also supposed to be shown as rotations of the X axis in the X-Y plane, but they kind of look like they are not. Anyway, they are shown as positive Z-direction vectors. I also tried to show if you rotate the X axis by \pi, it is the same as rotating the X axis by -\pi. For this reason the rotation represented vector\pi\hat{Z}, where\hat{Z} is the unit vector in the Z direction, is the same as the rotation represented by -\pi\hat{Z}. The (X,Y,Z) points at the end of these two \pi length vectors, (0,0,\pi) and (0,0,-\pi), are two of the antipodal points that are identified with each other in the exercise statement. Rotation can occur on any three dimensional axis, I mean not just cardinal ones. Therefore, the maximum rotation, \pi or -\pi, forms the radius of the sphere. Of course, you could rotate further than \pi, but that is the same as rotating in a negative direction. And visa-versa.

The three dimensional space representing rotation is strangely connected. It is twisted in higher dimensions so that if you move off outward in one direction away, after a while you find yourself moving inward toward the center again from the opposite direction. Even though this thing is twisted in higher dimensions, the "surface" of it is still three dimensional. I am going to try to show how this three dimensional space is twisted by taking a two dimensional slice of it, and showing how that is twisted.

Here is another diagram representing a slice of the rotation space sphere. It is the intersection of the Z-Y plane with the sphere. I have also drawn the X axis.

Image

The points on this circle all correspond to rotations that take the positive X axis and point it in the negative X axis direction. Point B, or \pi\hat{Z}, represents a rotation about the Z axis, which moves both the X and Y axis to point in their opposite directions, as I have tried to show in the little picture in the upper right corner. The point E, or \pi\hat{Y} rotates about the Y axis, moving the Z and X axis to point in their opposite directions, as I have tried to show in the lower right. Both points B and E will rotate the X direction to the -X direction, as will any other point on the circle. Generally, different points on the circle represent rotations that reverse the X axis but will orient the Y and Z axis differently. However, in the special case of antipodal points same orientation of all axis will result for both. For example, the half turn rotation represented by point E gives the same orientation as the opposite direction half turn rotation represented by -E. The same for B and -B, A and -A, C and -C. Each pair represent rotations of \pi or -\pi about a certain axis (A-A, B-B, etc.), and as such the rotations represented by each member of these pairs will leave the rotated frame in the same orientation as the other member of the pair.

This two dimensional space is actually twisted in such a way that if you move of away from the center in one direction, say the B direction, you soon find yourself moving toward the center from -B direction. I am going to show a few more drawings showing how this two dimensional space is really connected. First, I am going to "stretch out" points E and -E.

Image

So even though now "points" E and -E are now "edges", remember that they are really points. (I don't even know if this really makes sense)

More stretching and misshaping the surface

Image

Here I have taken to showing the -E edge in a dotted line, because next we will twist the surface,

Image

and bend the ends together to connect A to -A, B to -B, C to -C and E to -E

Image

My attempt at drawing a Möbius strip. Now, in your mind bend the strip so that edges E and -E come together, forming a a twisted impossible inner-tube and a loop out of E-E edge. Then shrink the E-E loop back down to a point. Got it? Neither do I. Now add up all these slices back into a 3-surface. Pretty twisted.

So, A and -A are actually the same point, same for B and -B etc.

OK, back to a flat circle. In the next series of images I try to show the attempt of deforming a closed loop representing a 2\pi rotation on a slice of the rotation space, a circle as explained above. On the left I have tried to show how a closed loop, going through B, representing a 2\pi rotation would look on this circle. The dotted line represents how the loop goes through the twisted space to arrive at the antipodal point. The next image over we are trying to deform the loop, moving it over on the top of the circle to point C, but note how the bottom moves over to -C as we do this, because really they are the same point. There is just nothing you can do with the loop to get rid of it.

Image

The argument is the same for the sphere shaped rotation space, but I can't draw that. The loop comes out of the sphere, and goes back in at the antipodal spot. No way to deform the loop to a point.

Now, I show how it works with two loops representing a 4\pi rotation.

Image

Here we have a loop that goes around twice, and we move one of the paths along the edge of the circle as we did before. We deform this path until we end up with it going in different direction of the other path, so the two paths cancel each other out.


16 Aug 2009, 03:12

Joined: 19 Oct 2009, 20:10
Posts: 2
Post Re: Exercise [12.17]
Hi,

Your solution is pretty interesting; it is much alike the one I was attempting myself to develop before I faced some issues. The analogy you make between the rotation ball and some ‘hyper-Möbius strip’ is enlightening I think to see where comes the ‘spinorial’ aspect of rotation.

However, at the end of your message, you try to apply this representation of the 3-dimensional rotation space to the problem 12.02 and that’s where I have some problems to follow you. In fact, that’s where I faced some issues, because I wasn’t able to see how the rotation ball I had imagined could help me visualize how a 4\pi-rotation is identical to no rotation. To be honest, I don’t see how your last paragraph justifies that a loop turning twice around the rotation space can be continuously shrinked down to a point.

This comes front the first end note of chapter 12, in which Penrose says that the ‘shrinkability’ should be formally understood as a homotopic deformation. Homotopic deformations, as Penrose explains, don’t allow two neighboring paths of opposite direction to cancel out, which is a characteristic reserved to homologous deformations. However, what we would like to demonstrate is that a ‘4\pi-rotation’ loop can be continuously shrinked down to a point, i.e. by mean of homotopic deformations, but your argument uses homologous deformations.

A short research on the Internet ( I know it’s cheating, but if I hadn’t seen the answer, I would have never been able to convince myself of the correctness of this argument, which still causes me trouble) tells us how we can solve problem 12.02 . The ‘4\pi-rotation’ loop looks like two straight lines going for example from the North Pole of the rotation ball down to the South Pole. Then, this loop can be ‘homotopically’ deformed in order to bring it to the surface of the rotation ball so it now looks like the half of a great circle joining the North Pole and the South Pole. However, since every point of the rotation sphere is identified with its antipodal point, the ‘4\pi-rotation’ loop is really homotopy equivalent to a whole great circle and not just one half of it. Finally, this great circle can be continuously deformed to become a point on the surface.

The only problem I have with this argument is that we implicitly assume that starting from a point of the rotation ball and going to another point of it following one path gives the same result as taking a completely different path to go to the same end point. However, it is clearly not the case for two general non-infinitesimal rotations \theta and \phi (about two different axis) applied successively on an object O ; the configuration ( \theta \phi ) O of the object at the end is generally totally different from ( \phi \theta ) O. In this sense, try for example to apply a rotation \frac {\pi}{2} in the direction of positive Y and then a rotation \frac {\pi}{2} in the direction of positive Z. The resultant point of the rotation ball, and hence the resulting orientation of an object in space, should be the same as having applied instead the rotation \frac {\pi}{2} \hat z followed by the rotation \frac {\pi}{2} \hat y . Unfortunately, the two resulting orientations are pretty different. So it is really not clear, at least for me, how we can say that a ‘4\pi-rotation’ loop is homotopy equivalent to a great circle on the rotation sphere on the basis that they are both paths joining the same two points of the rotation ball. I hope someone here would be able to explain me how it can be so.

Thanks,

Universus

NB : I'm really sorry for all the mistakes appearing in this message ; English is not my first language and I am still learning it.


22 Oct 2009, 19:24

Joined: 22 Apr 2010, 15:52
Posts: 43
Location: Olpe, Germany
Post Re: Exercise [12.17]
DimBulb wrote:
Here we have a loop that goes around twice, and we move one of the paths along the edge of the circle as we did before. We deform this path until we end up with it going in different direction of the other path, so the two paths cancel each other out.


"Universus" pointed out the following problem of this solution:

Universus wrote:
This comes front the first end note of chapter 12, in which Penrose says that the ‘shrinkability’ should be formally understood as a homotopic deformation. Homotopic deformations, as Penrose explains, don’t allow two neighboring paths of opposite direction to cancel out, which is a characteristic reserved to homologous deformations. However, what we would like to demonstrate is that a ‘4\pi-rotation’ loop can be continuously shrinked down to a point, i.e. by mean of homotopic deformations, but your argument uses homologous deformations.


I think Universus' objection is right, but that it is also easy to remedy DimBulb's solution: Instead of considering the CANCELLATION of two opposite paths, you just have to further continuously DEFORM the path such that it moves away from the center in the directions of points A and -A. DimBulb's last drawing can already be understood in this way. Thus the complete path shrinks to one point (as A and -A are the same point!).


12 May 2010, 18:09
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