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I am pretty sure I have studied every interpretation and explanation of spinors and the non-simply-connectedness of SO(3) rotations, and I eventually finally un
by ajkjk 6y ago
I am pretty sure I have studied every interpretation and explanation of spinors and the non-simply-connectedness of SO(3) rotations, and I eventually finally understood it -- but, by ignoring explanations like the 'belt trick', rather than embracing them. I don't think this article makes it any better. As long as you're stuck on quaternions you're not going to be able to see what's going on, and the belt metaphor just adds complexity as well.
The version that currently makes the most sense to me is this:
* Quaternions are a complete distraction and should be ignored. i, j, and k are (xy), (yz), (zx) bivector rotation operators (up to a factor of -i or something like that). Pauli matrices are the same and should also be ignored.
* The factors of "theta/2" in the exponents that are used in representations of rotations using quaternions (rotating vectors with e^(R θ/2) v e^(-R θ/2)) are distracting and should be ignored.
* The best way to see what is meant by "the space of rotations in SO(3) is not simply connected", you need to think about paths in _rotation_ space carefully. More carefully than I did as an undergrad! (Although I'm sure this is obvious to people who have studied the approach math in a course?)
The wrong approach -- which I was stuck on for years -- is to think about a point on a sphere in R^3 moving around. A vector that starts at, say, +z in R^3 and is rotated by 2pi in the xz plane ends up where it started exactly.
The key insight is that 'non-simply-connectedness' refers to _paths of rotations_, rather than paths of what the rotations act on. So imagine gradually rotating that vector +z in the (xz) plane. If you go around by 2pi, you've made a path that can be modeled as an exponential operation: e^(2pi (z^x)).
The question is: can this path be deformed to the identity path? It seems like it can -- you just change from rotating +z all the way around a great circle, to a smaller and smaller circular path until it is just rotating in place (and xy rotation). But somehow when you collapse it, you still have a 2pi rotation! It's just that now it's a 2pi rotation in xy rather than in zx. No matter what you do, if you collapse the rotation path to be the identity on the +z vector, the resulting path rotates _some_ vector by 2pi, instead of keeping it at the identity.
So in physics, it's not that there is a negative phase factor _per se_ that matters for physics. It's that there are two physically distinguishable states (identity rotations and anti-identity rotations) -- so two different electron wave functions can't be identified as the same electron, because they are a 2pi rotation apart just due to how they got there. And the fact that we _model_ this is as a negative sign is entirely an artifact of our obfuscating choices of mathematics for the situation.
- ogogmad 6y agoQuaternions enable one to formally prove that SO(3) is not simply connected. Here's how: The path in quaternion space e^{\pi i t} is not a loop, but its image in SO(3) under the covering map is a loop. Any contraction of the image loop would imply a contraction of its pre-image, which is impossible.
- ajkjk 6y agoyes, but that relies on knowing that the quaternions are such a cover, which is entirely non-obvious. An explanation that starts and ends with SO(3) without roaming into quaternions is necessary for intuition.
- moron4hire 6y agoI thought this was a better visualization, though he doesn't seem to mention you have to consider the whole system of the cup and his arm https://youtu.be/JDJKfs3HqRg https://youtu.be/JDJKfs3HqRg
- evanwise 6y agoAnother situation where geometric algebra makes everything clearer.
- ajkjk 6y agoin a sense, but I have some gripes with that too. It's bivectors that help a lot. The 'geometric product' I'm less sure about (this is always my complaint when GA is brought up).
- Koshkin 6y ago> i, j, and k are ... bivector rotation operators Well, that’s like calling the imaginary unit “a rotation operator.” (I mean, sure, you can do things like that with abstract algebraic constructs, marvel for a few moments, but then you move on, searching for the true meaning - which never happens to lie with any particular representation or an application.)
- 6y ago