Recent content by mathwonk

nice spoiler. took me a while to grasp, the key for me being that the map psi is itself bilinear, indeed it is the universal bilinear map.

absurd "coincidence": when I "solved " it (completely wrong) by confusing tangent with cosine, I got sqrt(11/4); and the actual answer it seems is roughly 11/4. I apparently enjoy numerical coincidences, even nonsensical ones.

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...

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

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...

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...

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...

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...

@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...

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...

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...

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...

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...

maybe not exactly research related, but I just want to suggest one always double check results announced by AI for accuracy. Here is an example: In some of my (joint) papers there are new methods introduced, not for themselves, but to solve a specific problem. Hence the paper and its title and...

My brother was practicing backstage with Bill Monroe at a bluegrass festival in 1967, to which Porter Wagoner had also been invited, when Dolly Parton, age 21, came up and said she really enjoyed his fiddling. She was quite attractive and his bandmates asked him how he knew her. He told them...