# I Set of possible rotations of a 3D object and the rotation history

1. Feb 11, 2019 at 7:26 PM

### Spinnor

I am trying to understand the picture below which is of a contractible and uncontractible loop in what I would call (proper name?) "rotation space", where "rotation space" is a solid ball of radius π with opposite points on the surface of the ball identified, each point of the ball representing a direction and angle less than or equal to π which could represent the rotation of an object. I understand what a point represents in such a space but I am having a hard time understanding what a path in such a space represents, for example the two paths below. I am thinking that the path represents some kind of "rotation history" of an object but something does not seem right about that idea. I know the answer must be simple but I am stuck.

The above was found at, http://www.damtp.cam.ac.uk/user/examples/D18S.pdf

(which Google Chrome tells me is not secure)

I think my second question is related so I ask it here. I would like to represent the rotational history of an object, say a soccer ball. Let the soccer ball have a spherical coordinate system printed on it. Say the ball is rotating relative to say a soccer field, we will(?) need 6 (edit, 7) coordinates to describe the rotational history of the ball, 2 spherical coordinates give the orientation of the ball with respect the the soccer field, (edit, another angle is needed to to fix the orientation), 2 more coordinates give the orientation of the instantaneous angular momentum vector L of the soccer ball, another coordinate gives the magnitude of L and one more coordinate for time. Is 6 (edit, 7) the minimum number of coordinates to describe the rotational history of the ball or are some of my coordinates redundant? Is such a space the same as S^2xS^2xR^2 (edit, S^2xS^3xR^2) ?

Edit, I should not post questions at the end of the day. If we know the orientation of the soccer ball as a function of time, 3 coordinates plus time, we know the rotational history of the soccer ball, right?

Edit, sorry for mistakes.

Thanks for any help.

Last edited: Feb 11, 2019 at 7:51 PM
2. Feb 12, 2019 at 4:14 AM

### haruspex

Yes, it's not clear.

If we pick a point in the ball it represents a rotation about such and such an axis through some angle. If move out along a radius of the ball that gets to a greaterrotation about the same axis.
But what if we move the point a little tangentially within the ball? That becomes an equal rotation but about a slightly different axis. This is like a precession, no? The path is then something like integration of spinors??

3. Feb 12, 2019 at 6:51 AM

### Spinnor

This is where my blockage is. If the angle does not change but the orientation of the rotation axis changes then an object can not have moved but that can still represent a path in our rotation space. An object that is precessing is constantly rotating so both the angle and orientation change at a constant rate? In our rotation space that path would be a double "conical spiral"? Still not where I want to be. Thanks.

4. Feb 12, 2019 at 12:02 PM

### lavinia

If the angle is zero then changing rotation axis gives a constant path $0⋅N(t)$ that stays put at the origin.

It might be helpful to formalize this a little bit:

Consider the space $S^2×[0,π]$ consisting of sphere of radius 1 Cartesian product the closed interval $[0,π]$. Any path in this space is of the form $(N(t),θ(t))$.

If one identifies all pairs $(N,0)$ to a point then the quotient space is homeomorphic to the 3 ball of radius $π$. The identification map is $(N,θ)→θ⋅N$.

A path of the form $(N(t),0)$ corresponds to changing the axis of rotation while keeping the sphere stationary. It becomes the path $0⋅N(t)$ in the 3 ball. This path is constant at the origin.

Aside: The identification mapping $S^2×[0,π]→B^3$ shows that the 3 ball is the topological cone on the 2 sphere. The cone on any topological space is the quotient space of its Cartesian product with an interval obtained by identifying one of the boundaries to a point. In this case the boundary sphere $S^2×{0}$ is identified to a point. This terminology generalizes the idea of the standard cone which is a cylinder with one of its boundary circles identified to a point. To see this, flatten the cone out by projecting it vertically onto a plane on which it is standing. The image is a circle with a spray of perpendicular radial lines that converge at its center. This is a disk. In general the $n+1$ ball is the cone on the $n$ sphere.

Last edited: Feb 12, 2019 at 4:48 PM
5. Feb 12, 2019 at 8:44 PM

### lavinia

Here is an example of a path of rotations through an angle of $π/2$ with rotation axis the line through $(cos(α),sin(α),0)$. Letting $α$ start at $0$ and end with $π/2$ the path starts with a rotation of $π/2$ around the $x$-axis and ends with a rotation of $π/2$ around the $y$-axis . This path traces a circular arc in the 3 ball.

$\begin{pmatrix} cos^2(α)&cos(α)sin(α)&sin(α)\\cos(α)sin(α)&sin^2(α)&-cos(α)\\-sin(α)&cos(α)&0\end{pmatrix}$

6. Feb 12, 2019 at 9:47 PM

### haruspex

By angle there I assume you mean the magnitude of the rotation of the object.
But I do not understand why you say the object "can not have moved".
Say we rotate the object through angle θ about some axis α. This is represented by a point in the ball distance θ from the origin and with direction corresponding to α.
If we move the representative point so that it keeps radius θ but is now in direction α', the corresponding motion of the object can be constructed as:
1. Rotate through -θ about axis α
2. Rotate through +θ about axis α'
Clearly this produces a net rotation of the object, and the smaller the difference between α and α' the smaller the net rotation.
For algebraic details see above post by @lavinia (but I think there is a sign error in one of the cos(α)sin(α) terms).
To follow the continuous path, apply the small angle approximation.

7. Feb 12, 2019 at 11:34 PM

### Spinnor

I was confused and didn't understand the little I thought I did (I think you are right, I'm wrong). The simplest of paths I think I understand, for example a line from the origin in the +z direction of length π/2 just represents the continuous rotation from 0 to π/2 about the z axis. It is the curved paths that are more complicated but now maybe I have it. So consider the path below in our rotation space, start at 0 then go to 1,2,3,4 and back to 1. That is, beginning at the point 1 we go in a circular orbit about the origin of radius π/2. So at this point it helps me to pick up a squarish empty plastic container of nuts and perform the rotations of my container given by points 1,2,3, and 4 and also points in between. So as I go around this rotation path my plastic container will smoothly rotate between the orientations 1,2,3,and 4 given below and when it gets back to point 1 it will have the same orientation as it had the first time it got to point 1?

Did I get the path in rotation space of a rotating top that processes in the Earth's gravitational field right? Both the angle and orientation change at constant rates giving some kind of conical spiral path in rotation space?

Thanks.

8. Feb 13, 2019 at 12:20 AM

### haruspex

Let's say rotations are positive clockwise, viewed from the ball's centre.
If in the cube diagrams the x axis is positive into the page and the y axis positive right, then yes.

9. Feb 13, 2019 at 7:59 PM

### lavinia

There is no sign error. One can check that the matrix has determinant 1 and that its transpose is its inverse.

A good visualization would be to draw the three vector orthogonal frame determined by the column vectors of the matrix and watch how it moves as the angle moves from $0$ to $π/2$.

Last edited: Feb 14, 2019 at 5:06 AM
10. Feb 13, 2019 at 8:03 PM

### haruspex

Thanks for checkng.

11. Feb 13, 2019 at 8:28 PM

### Spinnor

Is the following the path?

Thanks.

12. Feb 14, 2019 at 2:52 AM

### bolbteppa

The diagram is just crudely representing a 'continuous' sequence of rotations in 3-D space as a continuous path in an abstract parameter space, you can try to set up a better picture with the line through a sphere (anti-podal points) idea if necessary. The point of the diagram is really to show that paths through SO(3) can be connected but not simply connected. Your counting of degrees of freedom is off, seems like you are conflating the counting of DOF of rigid bodies with the number of parameters needed for a rotation in 3D (i.e. the dimension of SO(3)).

13. Feb 14, 2019 at 5:20 AM

### lavinia

On the 3 ball digram the curve lies on the sphere of radius $π/2$. Such a curve has the property that it is a path of rotations which all are rotations of an angle of $π/2$ from the initial position. Instead of rotating with a fixed axis, this path changes the axis keeping the angle of rotation fixed.

14. Feb 14, 2019 at 6:31 AM

### Spinnor

Thank you. This then,

15. Feb 14, 2019 at 6:39 AM

### Spinnor

I was off but now I think I get what the path represents and the important property that a 4π rotation is equivalent to no rotation.

As for mapping the rotational history of an object it seems there are two equivalent ways to do this, a path in "rotation space" or a path in "orientation space"?

Thanks.

16. Feb 14, 2019 at 6:39 AM

### lavinia

Right. Although now that I think abut it I did not show that the curve is actually an arc of a circle but only that is lies completely on a sphere of radius $π/2$ in the 3 ball.

17. Feb 14, 2019 at 1:03 PM

### lavinia

@Spinnor Do you see why the loop in the diagram is not contractible?

18. Feb 14, 2019 at 5:41 PM

### Spinnor

I thought I knew the answer to that and from the paper it says, "It is clear intuitively that the loop in figure 1b cannot be shrunk to a point while keeping its ends fixed—it is non-contractible." but I am not so sure now. When the paper says "ends" of the loop in figure 1b, are those the points at the origin?

Edit, rereading paper, "ends" of the loop are the points at the origin.

Thank you.

Last edited: Feb 14, 2019 at 6:35 PM
19. Feb 14, 2019 at 6:51 PM

### Spinnor

I think I see it now, what would be a good test question to determine if I do "see it"?

Thanks.

20. Feb 14, 2019 at 7:58 PM

### lavinia

The intuition is that any contraction to the origin would have to pull the two points on the boundary - at the angle of $π$ - off of the boundary and bring them into the interior of the ball. But since these two points are the same - they represent the same rotation - this would mean that the curve breaks and is no longer a loop.

This same intuitive reasoning can be used to show that if you go around the curve twice the it can be contracted.

BTW: Your book also assumes that the idea of a continuous path of rotations actually makes sense. It assumes that the 3 ball with antipodal points on its boundary sphere identified is topologically the same as the rotation group and is not just a parameter space. You might like to prove this for yourself.

Last edited: Feb 16, 2019 at 7:20 AM