Visualizing left action of SO(3) on itself

mma

The SO(3) group is topologically a 3-dimensional ball of radius [tex]\pi[/tex], if the opposite points on its surface are identified with each other. (the name of it is 3-dimensional projective space). The center of the ball represents the unit element e of the group. An arbitrary point g in the ball represents a rotation with axis g-e and with angle of ||e-g|| (thinking the ball as a part of the 3-dimensional euclidean space).

I am curious to know how looks the natural group left action in this ball. This would be completely described if we knew the curve t -> exp(tv)g for one arbitrarily selected v of so(3) and for each g of SO(3). How look these curves in the ball?
In the case of g=e (i.e. the center of the ball), this curve is a straight line passing from the center to a point of the surface which is identified with the opposite point and from this opposite point back to the origin. But I can't imagine, what curves we get if we take an arbitrary g point in the ball instead of the center.

Doodle Bob

It might help to represent the elements of SO(3) as unit quaternions.

mathwonk

or to read artins algebra book, chapter on group representations.

mma

Thank you both for the hints!

mma

According to Wikipedia, the "imaginary part" of the quaternion representation of a rotation is almost the same as what I told in my first post. The only difference is that the length of the representing vectors is the sine of the half angle of the rotation and not the angle itself.
The rotation with axis [tex]u[/tex] unit vector and angle [tex]\alpha[/tex] is represented by the vector [tex]u\alpha[/tex] in the original representation and [tex]u\sin{\alpha/2}[/tex] in the quaternion representation. In quaternion representation, the product of rotations [tex]u\sin{\frac{\alpha}{2}[/tex] and [tex]v\sin{\frac{\beta}{2}[/tex]
is
[tex]v\sin{\frac{\beta}{2}\cos{\frac{\alpha}{2} + u\sin{\frac{\alpha}{2}\cos{\frac{\beta}{2} + \sin{\frac{\alpha}{2}\sin{\frac{\beta}{2} u \times v [/tex].

For example, if [tex]u = (1,0,0)[/tex] and [tex]v = (\cos{\gamma}, \sin{\gamma}, 0)[/tex],

then their product is:


[tex](\cos{\gamma}\sin{\frac{\beta}{2}\cos{\frac{\alpha}{2} + \sin{\frac{\alpha}{2}\cos{\frac{\beta}{2},
\sin{\gamma}\sin{\frac{\beta}{2}\cos{\frac{\alpha}{2},
\sin{\frac{\alpha}{2}\sin{\frac{\beta}{2}\sin{\gamma})[/tex]

With this, it is easy to visualize for example the action of a rotation on a straight line passing through the centre of the ball.

Let this straight line be

[tex]t \mapsto (\sin{\frac{t}{2}},0,0) [/tex], i.e. the x axis of our coordinate system each point of it representing a rotation with angle [tex]t[/tex] arount the x axis.

Then the right action of the rotation around the axis [tex]v = (\cos{g}, \sin{g}, 0)[/tex] and angle [tex]b[/tex] transforms this straight line to the curve

x(t) = cos(g)*sin(b/2)*cos(t/2) + sin(t/2)*cos(b/2)

y(t) = sin(g)*sin(b/2)*cos(t/2)

z(t) = sin(t/2)*sin(b/2)*sin(g)

The left action of it only differs in the sign of z(t).


Giving explicit values to g and b, we can visualize these curves e.g. on the webpage http://cs.jsu.edu/mcis/faculty/leath...html#applettop

mma

Here is an example.

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