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 it is 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.

 

[lavinia]

The Tangent Circle Bundle of the 2-Sphere

Here is a way to “visualize” the tangent circle bundle of the 2-sphere that connects to a lot of rich topology.

Start with a smooth vector field, V, that has a single isolated zero at the north pole of the sphere. This vector field has index 2 which means that is winds around the the zero twice.

Divide each vector except the zero by its length to get a unit-length vector field. That is a field of vectors that lies in the unit tangent circle bundle except at the north pole. Now remove a small open disk around the north pole and let the radius of the disk shrink to zero. The unit-length vectors at the boundary of this disk form a closed curve that converges onto the fiber at the zero and in the limit attaches to it. One can imagine this boundary curve winding around the zero point getting closer and closer until it finally lands on the fiber above the zero by winding around it twice. For simplicity, assume a choice of vector field where this attaching map is strictly 2-to-1.

The complement of the removed open disk is a topological closed disk, D. The normalized vector field over it is also a closed disk because it is the image of a smooth section of the bundle over D. As the radius of the open disk shrinks to zero, this disk of unit vectors expands over the entire sphere, and its boundary circle attaches to the fiber above the north pole by the 2 to 1 mapping on its boundary circle . One then gets a closed 2-disk attached along its boundary by a continuous 2 to 1 mapping to a circle. This is a surface homeomorphic to the real projective plane RP^2.

Now, imagine rotating this normalized vector field through an angle. This rotated vector field produces a second projective plane that is also attached to the fiber circle at the north pole by a 2-to-1 mapping. Sweeping through all possible rotations gives a set of vector fields that span the entire unit tangent circle bundle and fill it with projective planes that intersect only in a common circle. Topologically this configuration is a circle of closed disks that are bound to a single circle each by a 2-to-1 mapping on their boundaries.

Alternatively, look back at the bundle over the closed disk, D, before the limit is taken. The set of all these rotated vector fields fills out a solid torus, D × S¹. Letting the radius of the disk shrink to zero, this solid torus expands and its boundary, a regular torus, T, becomes glued to the fiber at the north pole. T is completely covered by the parallel trajectories of the rotated vector fields. When attached to the fiber circle, the 2-to-1 mapping collapses each of these parallel trajectories onto the fiber.

So the unit tangent circle bundle can be described as a solid torus whose boundary torus, T, is attached to a circle by a mapping that collapses a family of parallel closed curves covering T by a 2 to 1 mapping on each curve.


Notes

The method of visualizing a topological space by breaking it into simpler pieces and then describing how they are put back together is a foundational technique in geometric topology.

The Poincaré–Hopf Index Theorem: On a compact smooth manifold, the sum of the indices of a vector field with isolated zeros must equal its Euler characteristic. Since χ(S²) = 2, a single isolated zero must have index 2. It is also true that on a connected closed manifold a vector field with an isolated zero can always be found no matter what its dimension.

Connections to Other Circle Bundles: Other circle bundles over the 2-sphere can be constructed using this exact geometric framework, altered only by changing the degree of the attaching map. In the case of the 3-sphere (corresponding to the Hopf fibration), the attaching map is 1-to-1. For other bundles , the mapping is n-to-1 with n > 2 and the individual "pages" are not manifolds. For the cases where the circle bundles are not tangent, the vector fields aren’t vector fields in the sense of flow fields on the sphere. They are not tangent to the sphere. Instead, they are called sections of the vector bundle.

 

[mathwonk]

All my comments derive from contemplating Lavinia's elucidating remarks. I have not digested them all, so I may make some mistakes, or repeat some insights.

Some ways to see that SO(3) is not simply connected.

1). We want to find a loop in SO(3) that does not shrink to a point. Just take any axis and consider the subgroup of all rotations about that axis. That subgroup is isomorphic, and homeomorphic, to the multiplicative circle group of complex numbers of unit length. This is not a proof, but no matter how I try, I cannot see how to shrink that circle of rotations to any smaller set of rotations.

It follows by the way immediately that SO(3) is a circle bundle over some space, since any group is a disjoint union of the cosets of any subgroup, and any coset of our circle subgroup is also homeomorphic to a circle. So SO(3) is a circle bundle over the coset space SO(3)/(our circle subgroup).

Moreover this coset space is just the 2-sphere S^2, as seen as follows: SO(3) acts transitively on the points of the 2 -sphere, i.e. every point can be taken to any other point by an appropriate rotation. Also, if we fix one special point p, then SO(3) is decomposed into cosets, by considering for each point q of S^2, the subset of rotations that take place to q. When p = q, this is just the rotation circle group of rotations about p, and for other values of q, it is the coset of this subgroup translated by any rotation taking p to q. Thus the map from SO(3) to S^2, taking each rotation to the image of p under that rotation, expresses SO(3) as a circle bundle over S^2.

 

[mathwonk]

2) Second, more sophisticated: the fact that SO(3) has a connected degree two covering space, namely SU(2), already implies that SO(3) is not simply connected. This uses the concept of “lifting” a path from a base space to a covering space.

I.e. fix a point p of the base space and a point q over it in the covering space. Then a path in the base space starting at p has exactly one path lying over it in the covering space and starting at q. This means that some paths starting at p in the base space which are “loops”, i.e. which end again at p, will have lifts which are not loops upstairs.

I.e. since the covering space is connected, we can join q by a path to the other point r also lying over p. Then the push down of this path to the base space will both begin and end at p, but the lift, which is unique, hence must be the original path, will begin at q and end at r.

Moreover, and this is the key fact, two paths in the base space which are homotopic with fixed endpoints, will have lifts that end at the same point of the covering space. Hence the loop in the base space which is the push down of the path upstairs joining q to r, is not homotopic to the constant path downstairs, whose only lift is the constant path beginning and ending at q.

Thus two paths which are homotopic in SO(3) will lift to paths in SU(2) ending at the same point over p. But conversely, two paths starting at p in SO(3) whose lifts in SU(2) begin and end at the same point will be homotopic in SO(3). I.e. since SU(2) ≈ S^3 is a sphere, hence simply connected, two paths beginning and ending at the same point of SU(2) must be homotopic, and then the homotopy pushes down to a homotopy of the two original paths in SO(3).

Hence the homotopy classes of loops in SO(3) starting and ending at our base point, are in one-one correspondence with the two points of SU(2) lying over that base point, i.e. the group of such homotopy classes in SO(3) has two elements hence is Z/2Z.

 

[Matterwave]

Thank you all for the comments and feedback!

They look very detailed. I will try to review as best I can later today. :)

 

[mathwonk]

3) We can see directly that SO(3) is topologically the 3 dimensional real projective space, which is not simply connected. This seems to follow from knowing it is obtained as a quotient of S^3 ≈ SU(2) by a faithful action of Z/2Z, i.e. the antipodal involution, but we can also see it as follows.

Note that every rotation of the 2-sphere is obtained by choosing an oriented axis and an angle of rotation. In particular, there is a surjection onto SO(3) from the product S^2 x S^1, where a point of S^2 determines an oriented axis, and an element of S^1 ≈ {complex numbers of length one} determines a counterclockwise angle. In fact since counterclockwise rotation through t radians about the oriented axis determined by the point p, equals the counterclockwise rotation through -t radians about then oriented axis determined by the opposite point -p, we have a surjection already from the product S^2 x [0,π], which is topologically a solid ball, but with an open concentric ball removed from the center.

Now this surjection is not yet injective, since the rotations with angle 0, about every axis, give the same identity rotation. So topologically we have to collapse the inner sphere to a point, now giving us topologically a solid closed ball of radius π, centered at the origin.

But we still have antipodal points of the surface of this solid ball yielding the same rotation, since opposite points p and -p determine the same axis, just with opposite orientations, and hence for each p the rotation determined by (p,π) is the same as the rotation determined by (-p,π). Hence finally SO(3) itself is obtained topologically by identifying antipodal points on the outer surface of this solid ball, i.e. this is just real projective 3 space.

Now we can see a loop that is not homotopic to a point by taking the image of any diameter of the solid ball; i.e. since the two antipodal points are identified, the diameter is now a loop in SO(3). And if we start at the center, i.e. the identity element of the group SO(3), and travel out along a radius until we reach the boundary point p on the outer sphere, then jump to the antipodal point -p, and continue in along that opposite radius to the center again, I claim this loop is exactly the rotation circle group about the axis determined by p.

I.e. from our construction, the value of the distance of a point of the ball from the center is the angle of the rotation, and for any point other than the center, the point of the boundary sphere at the end of the radius through the point, determines the axis. Hence our diameter determines exactly the circle group of rotations about the unique axis determined by that diameter.

Remark; All these answers are aimed at addressing the non simple connectedness of SO(3). The question of visualization of its group structure I think is somewhat different and builds for me off lavinia's discussions of its various actions on appropriate sets. But enough for now.

 

[mathwonk]

I just want to remark that if we think in terms of matrices, then it is obvious just from the definition that SO(3) is a circle bundle over the 2-sphere, and that in fact it is the unit tangent bundle over the 2-sphere.
I.e. by definition a 3x3 real matrix represents an element of SO(3) if and only if the columns are of length one, are mutually perpendicular, and define the right hand orientation of 3-space, i.e. the determinant of the matrix equals 1. This means that the third column is completely determined by the first two. Thus the group SO(3) is homeomorphic to the space of all ordered pairs of mutually perpendicular unit vectors in R^3.
Now the first column is any vector of length one in R^3, hence any point of S^2. If we just map such a matrix onto its first column, this defines a surjective map from SO(3) to S^2. The fiber over a point, i.e. over a given first column, is all possible choices for the second column. Since the second column can be any vector of length one perpendicular to the given first column, the possible choices fill out the circle of vectors of length one perpendicular to the first column vector.
It is also obvious that such vectors represent exactly the unit tangent vectors at the point of S^2 represented by the first column vector. I.e. a vector tangent to the sphere at p is a vector perpendicular to the radius vector at p, and since it lies in the trivial tangent bundle of 3 - space, we can translate it uniquely to the origin, placing its foot at the origin instead of at the point on the sphere, getting a unit vector perpendicular to our radius vector p. Thus the space of all ordered pairs of mutually perpendicular unit vectors in R^3, based at the origin, is homeomorphic both to the unit tangent bundle of S^2, and also to the group SO(3).
I.e. the group SO(3) acts faithfully and transitively on the set of ordered right handed orthonormal bases of R^3, where the image of the standard basis under a rotation, is just the set of columns of the matrix of that rotation. But the set of such orthonormal oriented bases is homeomorphic to the set of all ordered pairs of orthogonal unit vectors i.e. to the unit tangent bundle of S^2.

 

[Matterwave]

I spent a couple of hours reading, and up to now I have only digested 1 post haha. So, it will be a while before I can get to read the rest of the posts.

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.

Got it. Am I reading this right that you are picking out the specific circle bundle over the two sphere that $SO(3)$ must be diffeomorphic to as the tangent circle bundle? Am I right to assume that "tangent circle bundle" (over $S^2$) means taking only the unit vectors at each tangent space so that I can get the right dimensions out of it?

In other words, if I start with a generic circle bundle $(E, S^2, \pi, S^1, G)$ where $E$ is some total space, and $\pi$ an arbitrary trivialization, and $G=SO(2)$ is the structure group, then the correct circle bundle to have in mind when thinking about $SO(3)$ is $(UTS^2, S^2, \pi: (x, v) \rightarrow x, S^1, G)$ where $SO(3) \cong UTS^2$ and $UTS^2$ is the "tangent circle bundle" (of unit vectors)?

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.

Interesting visual. It reminds me of how we paste together a mobius strip vs pasting together the typical tangent bundle $TS^1$, but the pasting here seems much more complicated haha.

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.

This is also interesting!

 

[mathwonk]

@Matterwave:
The answer to your first part questions is yes, the specific circle bundle is the unit tangent bundle to S^2. But this can be made very explicit and concrete.
In post #14, I explain that an element of the unit tangent bundle to the 2-sphere S^2, is just an ordered pair of orthogonal unit vectors, e.g. the first two columns of a matrix in SO(3). Hence this homeomorphism of the unit tangent bundle with SO(3) is given by the map sending a matrix in SO(3) to its first two columns.

The circle bundle map SO(3)-->S^2, is then given by sending a matrix to just its first column, (since the first column is an arbitrary point of S^2).

In Lavinia's description, note that since a rotation is linear, and all tangent vectors can be taken as based at the origin, a rotation equals its own derivative. I.e. in her notation, if x,v is a pair of orthogonal unit vectors then the rotation r just acts by taking them to the pair (r(x), dr(v)) = (r(x), r(v)) of orthogonal unit vectors.

 

[wrobel]

A full $2\pi$ rotation of a rigid body about a fixed axis forms a non-contractible loop in $SO(3)$.

 

[wrobel]

see also https://en.wikipedia.org/wiki/Plate_trick

 

[lavinia]

I spent a couple of hours reading, and up to now I have only digested 1 post haha. So, it will be a while before I can get to read the rest of the posts.


Got it. Am I reading this right that you are picking out the specific circle bundle over the two sphere that $SO(3)$ must be diffeomorphic to as the tangent circle bundle? Am I right to assume that "tangent circle bundle" (over $S^2$) means taking only the unit vectors at each tangent space so that I can get the right dimensions out of it?

In other words, if I start with a generic circle bundle $(E, S^2, \pi, S^1, G)$ where $E$ is some total space, and $\pi$ an arbitrary trivialization, and $G=SO(2)$ is the structure group, then the correct circle bundle to have in mind when thinking about $SO(3)$ is $(UTS^2, S^2, \pi: (x, v) \rightarrow x, S^1, G)$ where $SO(3) \cong UTS^2$ and $UTS^2$ is the "tangent circle bundle" (of unit vectors)?


Interesting visual. It reminds me of how we paste together a mobius strip vs pasting together the typical tangent bundle $TS^1$, but the pasting here seems much more complicated haha.


This is also interesting!

yes. The tangent space to the sphere at any point is just a two dimensional plane so the vectors of length 1 in each plane make a circle centered at the origin. Geometrically, one can think of the tangent bundle as a subset of R^3xR^3 as all points (x,v) where x is a point on the sphere and v is tangent to the sphere at x. The unit circle bundle is all points where v is of length 1.

One might wonder if six dimensions is a little over kill to realize the bundle in Euclidean space since it is a 3 dimensional manifold. And it is but I think the best that can be done is 5 dimensions. 5 can be done using stereographic projection from the sphere in 6 dimensions into 5 dimensions after scaling the circle bundle to be of distance 1 to the origin.

 

[Matterwave]

see also https://en.wikipedia.org/wiki/Plate_trick

I have done the belt trick version of this with my belt! It was very illuminating :)

 

[Matterwave]

I.e. by definition a 3x3 real matrix represents an element of SO(3) if and only if the columns are of length one, are mutually perpendicular, and define the right hand orientation of 3-space, i.e. the determinant of the matrix equals 1. This means that the third column is completely determined by the first two. Thus the group SO(3) is homeomorphic to the space of all ordered pairs of mutually perpendicular unit vectors in R^3.

Yeah this makes sense! Let me try to write this out. You are describing thinking of a $SO(3)$ matrix like:
$$
O=\begin{pmatrix}\vert & \vert & \vert \\\mathbf{a} & \mathbf{b} & \mathbf{c} \\\vert & \vert & \vert\end{pmatrix}
$$

Where $||\mathbf{a}||=||\mathbf{b}||=||\mathbf{c}||=1$ and each pairwise dot product between $\mathbf{a},\mathbf{b},\mathbf{c}$ is equal to 0 and $\text{det}(O)=1$. Given $\mathbf{a}$ and $\mathbf{b}$, the final column $\mathbf{c}$ is fully specified.

Now the first column is any vector of length one in R^3, hence any point of S^2.

Right, $||\mathbf{a}||=1\rightarrow a_x^2+a_y^2+a_z^2=1$

If we just map such a matrix onto its first column, this defines a surjective map from SO(3) to S^2. The fiber over a point, i.e. over a given first column, is all possible choices for the second column. Since the second column can be any vector of length one perpendicular to the given first column, the possible choices fill out the circle of vectors of length one perpendicular to the first column vector.

Given $||\mathbf{b}||=1$ and $\mathbf{a}\cdot\mathbf{b}=0$ I can visualize the 2-D plane perpendicular to $\mathbf{a}$ and then restrict to the points on that plane which are unit distance from the origin (where the base of the vector $\mathbf{a}$ also lies).

It is also obvious that such vectors represent exactly the unit tangent vectors at the point of S^2 represented by the first column vector. I.e. a vector tangent to the sphere at p is a vector perpendicular to the radius vector at p, and since it lies in the trivial tangent bundle of 3 - space, we can translate it uniquely to the origin, placing its foot at the origin instead of at the point on the sphere, getting a unit vector perpendicular to our radius vector p.

Yeah I already did that translation (from tip to origin) implicitly in my head when I took $\mathbf{a}\cdot\mathbf{b}=0$ above.

Thus the space of all ordered pairs of mutually perpendicular unit vectors in R^3, based at the origin, is homeomorphic both to the unit tangent bundle of S^2, and also to the group SO(3).

Sounds right to me, given what was just constructed.

I.e. the group SO(3) acts faithfully and transitively on the set of ordered right handed orthonormal bases of R^3, where the image of the standard basis under a rotation, is just the set of columns of the matrix of that rotation.

So here you are pointing out $\mathbf{a} = O\hat{\mathbf{x}}$, $\mathbf{b} = O\hat{\mathbf{y}}$, $\mathbf{c} = O\hat{\mathbf{z}}$?

But the set of such orthonormal oriented bases is homeomorphic to the set of all ordered pairs of orthogonal unit vectors i.e. to the unit tangent bundle of S^2.

Nice! This was very helpful, thank you!

 

[mathwonk]

disclosure: I have not studied bundle theory and euler classes, and was not aware that circle bundles are often assumed to have a given orientation, hence that those on S^2 are classified by the integers. I assumed a circle bundle was just a map, locally a product, with all fibers homeomorphic to a circle, and did not assume a given action on that fiber by O(2). Thus I thought circle bundles on S^2 were classified by the fundamental group. It seems that circle bundles on S^2 have fundamental group Z/nZ for some n, and are homeomorphic iff their fundamental groups are the same. But it seems a circle bundle with fundamental group Z/nZ for n ≠ 0 has two orientations, with euler classes n or -n.

Still it is not always agreed which orientation should be given to a particular naive circle bundle. E.g. the famous Hopf fibration, a circle bundle map S^3—>S^2, may apparently be given euler class either 1 or -1, depending on convention. Due to its central significance as a sort of generator of all circle bundles, it is often preferred to assign to “the Hopf fibration” the orientation with class +1. But there is a standard construction of this fibration as a sub-bundle of a complex line bundle which has chern class -1, hence apparently also the euler class is -1.

I.e. S^2 is famously homeomorphic to the complex projective line, i.e. the “Riemann sphere” CP^1. The basic construction of this space CP^1 is as the space of all complex lines through the origin of C^2 ≈ R^4. The unit sphere in that space is of course homeomorphic to S^3, and each complex “line” through the origin is a real plane, hence meets that sphere in a circle. The map sending a unit vector in C^2 to the complex line it spans in C^2, thus defines a map S^3—>CP^1 ≈ S^2, the Hopf fibration.

Since this complex line bundle is the “tautological” line bundle O(-1) over CP^1, i.e. the line over each “point” of CP^1 is just the complex line representing that point, and it is well known that this line bundle has chern class -1. In particular, it has no non - zero holomorphic sections. (Its dual bundle is the complex line bundle O(1) whose non trivial sections are non zero complex linear homogeneous polynomials, hence each having a single zero.) Thus apparently the euler class is also -1 with this orientation. In particular, there is only one map S^3—>S^2 with circles as fibers, the Hopf fibration, but its euler class is apparently either 1 or -1, depending on your choice of orientation of the bundle.

Next, I will try to explain how this one Hopf circle bundle in some sense generates all non trivial circle bundles over S^2.

 

[mathwonk]

Briefly, starting from the Hopf fibration, in each fiber just identify each point on the circle with its opposite point. Since identifying antipodal points on a circle yields again a circle, the quotient space is again a circle bundle over S^2, equipped with a map from the hopf bundle which is 2-1. Since the hopf bundle is simply connected, and maps 2-1 onto the new bundle, that new bundle has fundamental group Z/2Z. Hence that bundle is apparently homeomorphic to SO(3).
Similarly, one can construct from the Hopf bundle, by identifying sets of n points, each located 2π/n from the previous one, another circle bundle on S^2 with fundamental group Z/nZ. Since this gives all circle bundles on S^2, the Hopf bundle "generates" all of them. (Except the trivial one S^2 x S^1 with fund group Z.)

 

[wrobel]

By the way, here is one of the standard conclusions derived from the non-triviality of the configuration manifold (in this case, SO(3)).

It is well known that on a closed Riemannian manifold, every homotopy class of non-contractible closed curves contains a geodesic. Consequently, no matter what potential forces are applied to a rigid body with a fixed point, such a rigid body is bound to have a whole bunch of periodic motions.