Best visualization of SO(3)

[mathwonk]

Building again on lavinia's posts, we know SO(3) is homeomorphic to RP^3, the real projective 3-space, which is a solid ball with antipodal points on each diameter identified. She has described how to see this as a union of a family of projective planes sharing a common circle, i.e. of a P^1 of projective planes, sharing a common P^1.
It is easy to see that P^3 has such a decomposition just from projective geometry. I.e. just take any (projective) line in projective 3-space and consider all (projective) planes containing that line. This sweeps out P^3 as a union of copies of P^2 having a common line, i.e. a common P^1 or circle.
P^2 also has a decomposition as a union of projective lines with a common point. Just take any point in your P^2 and consider all lines through that point. This P^1 of lines with a common point sweeps out P^2.
One can also see these decompositions directly in the group SO(3). I.e. given a sphere, fix an axis, say joining the north and south poles. Now choose any plane through that axis, and consider all rotations of the sphere with axis in that plane. This subset of SO(3) is naturally a copy of P^2. E.g. this set of rotations has a natural decomposition as a union of copies of P^1 each with a common point. I.e. for each axis in this plane, the rotations about such an axis forms a P^1, and they all share the identity rotation, so rotations with axis in a given plane, are a union of a circular family of P^1's, all with a common point. As we vary the given plane, we decompose SO(3) as the union of a P^1 of copies of P^2, all with a common P^1 or circle.
Indeed one can see directly the structure of SO(3) as P^3 as follows: we know P^3 is obtained from a solid ball of radius π by identifying antipodal points on each diameter. This space is directly homeomorphic to SO(3) as follows: the center of the ball is the identity rotation. On each radius, the point at distance t from the center represents the counterclockwise rotation about the head of that radius through t radians. Since the counterclockwise rotation about the head of radius vector v through π radians, is the same as the counterclockwise rotation about the head of radius -v through π radians, SO(3) is obtained in this way from the solid ball, by identifying antipodal points on each diameter, i.e. SO(3) is thus naturally identified with RP^3.
As above, the points lying on a given plane containing the north and south poles correspond to rotations with axis in that plane, hence to a copy of P^2, and SO(3) is the union of a "pencil" of, i.e. a P^1 of, these copies of P^2, parametrized by the P^1 of planes through the north-south axis.

 

[mathwonk]

This is getting clearer, if slowly. We know every (non trivial) rotation about a given axis, belongs to a circle subgroup of rotations with that same axis, and that subgroup is homeomorphic to a circle. The subgroups of rotations about two different axes share only the identity rotation. So SO(3) is naturally a union of circles that all share one common point, the identity rotation. We just saw how to view SO(3) as a solid ball of radius π, with antipodal points of diameters identified, and now we can see that it is a union of circles representing these rotation subgroups with a common axis.
Namely the solid ball is the union of its diameters, and each diameter represents a circle (after identifying antipodal endpoints). This exhibits SO(3) as the union of these circular rotation subgroups, all having as common point the identity rotation (the center of the ball).
Apologies for possibly being the last guy to see how simple this is. I have really enjoyed thinking this through, guided by the answers here. But for me this picture completely illustrates the geometry/topology of SO(3). (It also shows the algebra at least of rotations about a given axis, as translations along a radius. It remains, for me, to see how to compose rotations about different axes in this picture.)

 

[lavinia]

This exhibits SO(3) as the union of these circular rotation subgroups, all having as common point the identity rotation (the center of the ball).

In terms of the sequence, SO(2)->SO(3)->S^2, since SO(2) is closed in SO(3) ,this sequence describes a fiber bundle whose fibers are cosets of some fixed copy of SO(2).

I guess from this approach one next asks which circle bundle over S^2 this is.

Under the action of SO(3) on the 2 sphere's tangent circle bundle ,a full rotation around an axis sweeps out a tangent fiber circle. So the diffeomorphism takes each coset to a different tangent circle and this decides the question.

 

[mathwonk]

RE: post #33: yes, I tried to say something like this in post #10.

 

[mathwonk]

I want to describe some geometry that seems to connect to lavinia's vector field discussion, which I have yet to grasp. Lavinia has shown how to view SO(3) ≈ RP^3 as a stack of copies of RP^2 with a circle in common. It helps therefore to think about the topology of RP^2, which can be viewed analogously as a stack of circles, or RP^1's, with a point in common.
Abstractly these decompositions are easy. In RP^3 just fix a line, and sweep out RP^3 by the family of planes through that line. If we choose any other line (P^1) not meeting the first line, there is exactly one such plane for each point on that other line, hence we have a P^1 of planes sweeping out RP^3, and each having a line (or circle) in common. In any RP^2, choosing a point and a line not meeting that point, we can sweep out RP^1 as a P^1 of lines through that point, each meeting our chosen line once.

In our case SO(3) ≈ RP^3 can be seen as a union of a P^1 of RP^2's, each with a circle in common as follows: Choose a north - south axis on the 2-sphere, and a geodesic through those poles. Then the subset of rotations of the sphere with axes joining antipodes of that geodesic, form an RP^2 inside the full RP^3 ≈ SO(3). SO(3) is swept out by the P^1 of such RP^2's corresponding to all choices of geodesics, i.e.planes, through the north and south poles.
The RP^2 of rotations with axes on our geodesic is itself visibly the union of a P^1 of circles, each with a point in common, since for each such axis, the rotations about that axis form a circle, and this P^1 of circles all have the identity rotation in common.
In addition to viewing RP^2 as a union of circles, each with common point, there are several other ways to view the topology of RP^2, each with its advantages. By definition, RP^2 is the result of identifying pairs of antipodal points of the 2-sphere. If we reduce this 2-1 map from the sphere to RP^2 to just the closed upper hemi-sphere, a closed disc, we see we get RP^2 just by identifying pairs of antipodal points on the circumference of a closed disc. But we also know that if we had gone further and identified the entire circumference of the closed disc to one point, we would obtain the 2-sphere again, topologically. This collapsing map induces a collapsing map also from RP^2 to the 2-sphere, collapsing one circle in RP^2 to a point. Thus not only is there a 2-1 map from the 2-sphere S^2 to RP^2, there is also a map from RP^2 to S^2 that is injective everywhere except for collapsing one circle in RP^2 to a point.
This map RP^2-->S^2 is called "blowing down". The inverse construction can be used to construct RP^2 out of S^2, by blowing up, i.e. by removing one point of S^2 and replacing it by a P^1 ≈ circle.
Topologically, removing one point of S^2 leaves the same result (i.e. an open disc) as removing a closed disc, so remove a closed disc from S^2 and then add in a boundary circle and then identify opposite points of that boundary circle. This replaces one point of S^2 by a P^1 ≈ circle. In other language, we attach a circle to the sphere minus an open disc, which is just a closed disc, by a 2-1 map from the boundary circle of the disc to the added circle. This is sometimes called adding a cross cap, and can also be achieved by sewing in a mobius strip onto the closed disc, by identifying the circular boundary of the mobius strip with the boundary circle of the closed disc. The result of blowing up the sphere at a point is to add in a new point for each tangent line through that point. A path on the 2-sphere that passes through the point p, and tangent to the vector v at p, yields a path on the blowup of p, that passes through the new point corresponding to the direction v. In particular two paths through the north pole of the 2-sphere with different tangent directions, no longer meet on the blowup.

OK, I am getting to the connection with lavinia's vector fields. Now let's try to represent the RP^2 of rotations with axes on a given geodesic, in these terms. To "see" a given rotation, look at its action on the North Pole, i.e. let the North Pole be p, and look at the orbit of p in S^2, under the action of a rotation whose axis lies in the plane of our given geodesic. This orbit is a circle passing through p and perpendicular to the given geodesic. As the axis rotates around the center of the sphere, remaining in the plane of our geodesic, these circles sweep out S^2, but all contain the North Pole p. Thus we see our S^2, (not RP^2), as a P^1 of circles, each with a common point; moreover note these circles are all tangent at p, (since all are perpendicular to the given geodesic through p). (These circles look like the flows of lavinia's index 2 vector field!)
But these circles only represent faithfully the rotations whose axis does not equal the north-south pole, i.e. does not contain p. That one circle group of rotations leaves p fixed. Thus our map sending a rotation with axis in our chosen plane to the image of p under that rotation, blows down the rotation group whose axis contains p, to a point. I claim this is the blowing down map from our RP^2 of rotations with axes in our given plane, to the 2-sphere. I.e. we can try to recover this RP^2 and hence our full family of rotations by blowing up p. I.e. since blowing up puts in all directions at p, and since rotations about p do act on directions, ... ooops!, the rotations around this axis do not act fact faithfully, but only 2-1 on the tangent lines at this point. I.e. each tangent line is fixed both by the identity rotation and the 180 degree rotation. So blowing up this point only seems to give me the quotient of my rotation group by that isotropy subgroup of 2 elements. what gives??
..well, like all these speculations, this is naive. for one thing the other rotations, about other axes, do not seem to act on the directions at p.....(but I really liked it at first.)