Best visualization of SO(3)

[Matterwave]

TL;DR

Trying to find a good visualization of SO(3)

I know a decent number of properties of SO(3) and obviously I can visualize the group action on three space as rotations. But a good visualization of the group itself eludes me.

SU(2) is somehow more intuitive to visualize since it simply is $S^3$. So I just think of a 2-d sphere and say "well it has one more dimension" (lol).

Sometimes, because SU(2) double covers SO(3) I see some descriptions of SO(3) as "half of $S^3$". Sometimes this is tempting but it has some undesirable properties. For one, imagining a sphere cut in half introduces an edge to the sphere. The cut is also entirely arbitrary.

Really, it's two antipodal points of SU(2) maps to one point on SO(3) but this doesn't help me "get a picture" in my head. At least not in a way where, for example, the non simply-connectedness of SO(3) becomes obvious to me.

Anyone know of some good visualizations? To help build intuition?

I suspect this will also help me understand non simply connected as more than "there's a hole".

 

[anuttarasammyak]

How about Bloch sphere for qubit? Perhaps you want more than that.

 

[wrobel]

To me, the best way to visualize $\text{SO}(3)$ is as a rigid body with a fixed point. In particular by examining its motion through Euler angles, it becomes clear that $\text{SO}(3)$ is a fiber bundle with base $\mathbb S^2$ and fiber $\mathbb S^1$.

 

[Paul Colby]

Visualize as a manifold seems to be all you consider. There’s the algebraic aspect of $SO(3)$ as well.

 

[lavinia]

To expand @wroblel's point in post #3: The action of SO(3) on the 2 sphere can be extended to an action on its tangent circle bundle by the mapping r:(x.v)-> (r(x),dr(v)) where r is a rotation of the sphere. This mapping defines a diffeomorphism of SO(3) onto the tangent circle bundle of the 2 sphere.

There are infinitely many different circle bundles over the 2 sphere. So one needs a way to distinguish them.

There are a few ways to do this. One is to notice that if one cuts any circle bundle along the equator of the sphere, it splits into two trivial circle bundles over the two hemishperes and since a hemisphere is homeomorphic to a closed disk, one gets two copies of D^2xS^1 , the Cartesian product of a closed disk with a circle and these are both topological solid tori. To see this,think of the Cartesian product as a circle of disks. So one sees that any circle bundle over the 2 sphere is two solid tori pasted together along their boundaries. Each is distinguised by the ways these pastings are done.

Interestingly, since the 3 sphere is the total space of the Hopf fibration which is itself a circle bundle over the 2 sphere , the sphere in four dimensions can be made from two solid tori that are pasted along their boundaries. To see this visually one one might try to see how this pasting happens through the stereographic projection of the 3 sphere into R^3 and then look at the way the images of the Clifford tori fit together.

There are other ways to do this which I am happy to describe.

 

[lavinia]

A mathematician showed me another way to see the 3 sphere.

Slice the 3 sphere along its equatorial 2 sphere. Its two hemispheres are 3 dimensional balls which can be seen by projecting them vertically into three dimensions. This is the same as slicing a regular sphere in three dimensions. If one imagines that the sphere is an inflated balloon, then the two hemispheres will naturally deflate until they become flat disks. Then they are two dimensional rather than three. So think of the three sphere as an inflated three dimensional balloon.

Next core out a solid tube from each of these balls . Make sure to take care that back in the 3 sphere, these tubes connect at the ends to form a solid torus. Put them together to make a solid torus. Now look at one of the cored solid balls. Imagine that is it very stretchable, and put your hands in the top of the hole an warp it into a wider circle and then pull it down until it is flush with the bottom of the hole and is concentric with it. In this process, warp the entire outside surface of the cored ball downward until it becomes flattened out. Note that the inner surface where the tube was removed will curve over during the stretching. This makes half of a solid torus, a half bagel. Do this with the other one and paste the two together to make a whole bagel. Now there are two solid tori.

One knows that the solid torus made from the two cores fits into the bagel through its hole but because of the strectching, in order to recover the spots where it came from, its boundary must be spread out over the entire boundary of the bagel. In other words the two solid tori are pasted together along their boundaries.

If I knew how to post pictures I could make some drawings but this visualization is a good one to struggle with one one's own.

 

[Ibix]

If I knew how to post pictures I could make some drawings but this visualization is a good one to struggle with one one's own.

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[Hornbein]

Slice the 3 sphere along its equatorial 2 sphere.

The equation of the equator of a 3-sphere is w^2 + x^2 = y^2 + z^2 = 1/2. It looks like the equation for a 2-sphere but unlike a 2-sphere it can't be embedded in a 3D space. Instead it is a sort of 2D torus embedded in 4D.

The equator of a 3-sphere is the set of points equidistant from the two "poles." These "poles" are both circles, w^2 + x^2 = 1 and y^2 + z^2 = 1. Every point on each circle is the same distance from every point on the other circle.

Each circle may be embedded in a 2D plane. The intersection of the two planes is the point at the center of the 3-sphere.