# Visualizing left action of SO(3) on itself

• mma
In summary, the SO(3) group is topologically a 3-dimensional ball of radius \pi, if the opposite points on its surface are identified with each other and the center represents the unit element e of the group. The natural group left action in this ball is described by the curve t -> exp(tv)g for one arbitrarily selected v of so(3) and for each g of SO(3). 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, when g=e. When an arbitrary point g is taken instead of the center, the resulting curve is given by x(t) = cos(g)*sin(b/

#### mma

The SO(3) group is topologically a 3-dimensional ball of radius $$\pi$$, 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.

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

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

Thank you both for the hints!

According to http://en.wikipedia.org/wiki/Quaternions_and_spatial_rotation" [Broken], 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 $$u$$ unit vector and angle $$\alpha$$ is represented by the vector $$u\alpha$$ in the original representation and $$u\sin{\alpha/2}$$ in the quaternion representation. In quaternion representation, the product of rotations $$u\sin{\frac{\alpha}{2}$$ and $$v\sin{\frac{\beta}{2}$$
is
$$v\sin{\frac{\beta}{2}\cos{\frac{\alpha}{2} + u\sin{\frac{\alpha}{2}\cos{\frac{\beta}{2} + \sin{\frac{\alpha}{2}\sin{\frac{\beta}{2} u \times v$$.

For example, if $$u = (1,0,0)$$ and $$v = (\cos{\gamma}, \sin{\gamma}, 0)$$,

then their product is:$$(\cos{\gamma}\sin{\frac{\beta}{2}\cos{\frac{\alpha}{2} + \sin{\frac{\alpha}{2}\cos{\frac{\beta}{2}, \sin{\gamma}\sin{\frac{\beta}{2}\cos{\frac{\alpha}{2}, \sin{\frac{\alpha}{2}\sin{\frac{\beta}{2}\sin{\gamma})$$

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

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

Then the right action of the rotation around the axis $$v = (\cos{g}, \sin{g}, 0)$$ and angle $$b$$ 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/leathrum/Mathlets/parapath.html#applettop"

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Here is an example.

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## 1. What is the purpose of visualizing left action of SO(3) on itself?

The purpose of visualizing left action of SO(3) on itself is to understand the geometric representation of rotations in three-dimensional space. SO(3) is the special orthogonal group of three-dimensional rotations, and by visualizing its left action on itself, we can see how rotations affect points and shapes in space.

## 2. How does the left action of SO(3) on itself work?

The left action of SO(3) on itself involves applying rotations from the special orthogonal group to points in three-dimensional space. This means that for each rotation in SO(3), there is a corresponding transformation of points in space. This left action can be visualized by using geometric tools such as matrices or quaternions.

## 3. What are some practical applications of visualizing left action of SO(3) on itself?

Visualizing left action of SO(3) on itself has practical applications in computer graphics, robotics, and physics. In computer graphics, it is used to create realistic 3D animations and simulations. In robotics, it is used to model and control the movement of robotic arms. In physics, it is used to understand the behavior of rigid bodies and their rotational motion.

## 4. How does visualizing left action of SO(3) on itself relate to linear algebra?

The left action of SO(3) on itself can be represented using matrices, which are an important tool in linear algebra. This allows us to use linear algebra techniques, such as matrix multiplication and eigendecomposition, to analyze and understand the left action of SO(3) on itself. It also helps us to see the connection between rotations in three-dimensional space and linear transformations.

## 5. Are there any limitations to visualizing left action of SO(3) on itself?

One limitation of visualizing left action of SO(3) on itself is that it only applies to rotations in three-dimensional space. It does not take into account other types of transformations or rotations in higher dimensions. Additionally, visualizing the left action may become difficult for complex or non-standard rotations, making it less useful in those cases.