0
$\begingroup$

I'm reading and trying to understand the following text, which leads to the conclusion of identifying the sphere $S^2$ with $SO(3) / SO(2)$:

enter image description here

So, with XYX parametrization we are trying to find three angles $\alpha$, $\beta$, $\gamma$ such that a a rotation $R_{\alpha, \beta, \gamma}$ can be expressed as a rotation of $\alpha$ over the $x$-axis, a rotation of $\beta$ over the $y$-axis and a rotation of $\gamma$ (again) over $x$-axis.

With Euler parametrization it looks to me that we are holding fixed the $x$ but allowing two angles rotation around it (?).

Can we really reach any point in the sphere rotating like this? I cannot imagine it really..

$\gamma$ is set to $0$ in above parametrization, 'but it should be clear that with any $\gamma$ the same point $\textbf{n}_{\alpha,\gamma}$ is reached.

I just don't understand it..

$\endgroup$
6
  • 1
    $\begingroup$ If you take a globe and place a two markers you can get from one to the other by first traveling along a line of latitude then traveling on a line of longitude. These are the two rotations on the sphere that allow you to travel to any point. Note that we have a tacitly fixed NS axis on a globe. $\endgroup$ Jan 19, 2022 at 13:25
  • $\begingroup$ thanks, so basically two rotations allow to reach any point on the sphere, i.e. we don't need any third angle $\gamma$.. is this correct? What do you mean exactly with note that we have a tacitly fixed NS axis on a globe'? $\endgroup$ Jan 19, 2022 at 13:42
  • $\begingroup$ Also why do we use the $XYX$ parametrization if we can accomplish any general rotation with two rotations? $\endgroup$ Jan 19, 2022 at 13:43
  • $\begingroup$ I only point out the fixed NS axis because we must have some vector that the angles are in reference to. Tacitly means unstated since we don't typically talk about this axis. As for your second question you can't accomplish any rotation, you can merely act transitively on the sphere. This is because many rotations can take one point to another. For example consider the direct route and then consider rotating by a half turn around the midpoint between the points. Each will leave the sphere is a much different configuration. Many of these are impossible with our generating set. $\endgroup$ Jan 19, 2022 at 14:20
  • 1
    $\begingroup$ Sure I'll write one up now. $\endgroup$ Jan 19, 2022 at 16:22

1 Answer 1

1
$\begingroup$

So firstly we have the $XYX$ parameterization of $SO(3)$ using three rotations which is different than the Euler angles that parameterize points on the sphere. The set of rotations is larger and we'll examine this in more detail.

Firstly, for the Euler angle parameterization we can use the analogy of a globe and travel alone lines of latitude and longitude to get from one place to another. Note that these rotations are defined by two great circles, one around the equator and one through the poles. Since this is a composition of two rotations it will again be a rotation because $SO(3)$ is closed and by Euler's rotation theorem we know that this rotation has a single fixed axis. This is because we chose special circles to rotate through.

However there are many other elements of $SO(3)$ that would take you from one point to another. Given two points $a$ and $b$ on the sphere they define a great circle that connects them directly and the plane that great circle lays in is orthogonal to a line. If we rotate about that line it will take $a$ directly to $b$ or vice versa in the shortest path along the sphere. This is equivalent to rotating the plane they're in.

By contrast we can take the stabilized axis to be the line that intersects the sphere at the midpoint between $a$ and $b$ and goes through the center. This rotation will flip the plane about the perpendicular bisector of the line connecting $a$ to $b$. Think if a piece of paper with a pencil glued to the middle then twiddling the pencil.

As and explicit example lets consider a short trip from one side of the north pole with the north pole as the midpoint. If I walk directly to the destination I will cross the north pole. Alternatively, I could walk in a circle which is centered at the north pole and once I'm half way around I will arrive at my destination. The first stabilizes some line on the equator and the second stabilizes the north pole.

In fact, there are infinitely many rotations that work and the midpoint is just a special case. For the points $a$ and $b$ there will be a great circle through the midpoint and orthogonal to the great circle connecting $a$ and $b$. I can choose any of the points on the midpoint great circle and use it to define an axis of rotation that will take $a$ to $b$. I would encourage you to investigate which axis minimizes the angle of rotation.

One useful definition is given a group $G$ that acts on a set $S$ the group is said to act transitively on $S$ if for all $s_1 ,s_2 \in S$ there exists a $g \in G$ such that $g \cdot s_1=s_2$. You can see where this definition comes from because if we want to transit from one point on the sphere to another using an action in $G$ we would need $G$ to have this property otherwise a route may not exist. The Orbit-Stabilizer Theorem is named after this example with an orbit stabilizing its axis of rotation. This will be relevant when examining the quotient.

$\endgroup$
1
  • $\begingroup$ what a detailed answer! I'll go through this step by step, MANY thanks! $\endgroup$ Jan 19, 2022 at 17:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .