PhoenixMercurous
Soaring through the stars
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I'm not sure I understand your question; the notation is standard in mathematics and not physics-specific. O(n) is the orthogonal group, the symmetry group of n-dimensional Euclidean space that preserves a fixed point, i.e. without translations. SO(n) is the specifial orthogonal group, i.e. also excluding mirror reflections, and thus is just the rotation in Euclidean n-space.
You can represent O(n) as real orthogonal matrices if you like, RTR=RRT = I, which have determinant ±1, and SO(n) would correspond to those with det R = +1. U(n) is the unitary group, which you can represent as complex unitary matrices, with analogous restriction so SU(n).
For example, SO(3) is the groups of rotation in three dimensions (i.e. symmetry of a 2-sphere). SU(2) is isomorphic to the electron spin group; the usual Pauli matrices are just a standard basis for it. If you take them and multiply by i, then including the identity matrix, you have an algebra isomorphic to the quaternions, which represent the same rotation by two different quaternions. So it's at least intuitive that there's a double cover. Here we do have some notional convention differences between mathematicians and physicists, but I gather that's not what you're asking about.
Anyway, one way to think about it is this: if you think your electron has a conserved angular momentum by itself, then guided by the insight of Noether's theorem, the conserved quantity should correspond to infinitesimal rotations. The infinitesimal limit of the rotation group SO(3) is its Lie group so(3). The notation is not important, but something special happens: the Lie group of SU(2) is isomorphic to so(3). So the requirement of angular momentum conservation can't distinguish between them: the spin 1/2 group must be a valid possibility.
There's a whole bunch of things that can be said in less handwavy way, and have been done in various textbooks and literature, up to and including that field theory essentially narrows the elementary particle spins to {0,1/2,1,3/2,2}, but I hope this clarified some things. The overall point is that characterising spin as some leftover bit and to imply this is some kind of problem or an ad hoc appendix is just wrong. Spin is there because the electron is spinning. (People that have a problem with this statement, and many do, are stuck in a classical mindset. Quantum-mechanically, to say that a physical thing does or doesn't do something is to at least implicitly make an empirical prediction about a result of some observable. Since an electron has intrinsic angular momentum we can measure, it is spinning.)
Also, that the Kepler problem has symmetry SO(4) is an interesting property of bound Newtonian orbits; the motion of the planet in momentum space corresponds through stereographic projection to motion of a free particle on a sphere in four dimensions. This fact was used to first derive the correct energy levels of hydrogen in 1925, i.e. a year before the Schrödinger equation.
With all due respect, you aren't very good at explaining things to non-physicists. The notation might be standard in pure mathematics and physics, but I've never seen it before, even in the masters level solid state physics course I took several years ago. I'm used to seeing the various symmetry notations used in crystallography, which are entirely different. Your digression into electron angular momentum is entirely irrelevant to my question and doesn't help clarify anything.
I looked up Lie groups, and lets just say I know just enough math to realize the phrase "differentiable manifold" means I'm way out of my depth. Reading a bit more, it seems the unfamiliar notation is because you're dealing with continuous symmetries while I've only dealt with discrete symmetries before. It makes sense that two different categories of symmetries would have different notations.
Anyways, I got enough of an answer to my question.