How Does U(1) Double-Cover SO(2) for a Specified Angle?

In summary, the relationship between U(1) and SO(2) can be understood by recognizing that U(1) represents the group of complex numbers with a modulus of 1, while SO(2) represents rotations in a two-dimensional plane. The concept of double-covering refers to the way U(1) maps onto SO(2) such that each rotation in SO(2) corresponds to two elements in U(1). Specifically, for a specified angle θ, the rotation in SO(2) can be represented by the complex exponential e^(iθ) in U(1), demonstrating that each angle θ has a corresponding angle θ + 2π, thus illustrating the double-covering nature of U(1) over SO
  • #1
pellis
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TL;DR Summary
Can someone please provide an explicit example of two complex numbers for double cover U(1) of SO(2) for a specified angle R(θ)?
I'm trying to find an explicit example showing exactly how the U(1) “circle group” of complex numbers double-covers 2D planar rotations R(θ) that form the rotation group SO(2).

There are various explanations available online, some of which are clear but seem to be at variance with other explanations. (I leave aside other more technical explanations suited only to graduate students in mathematics - I'm only a chemist - such as those in https://ncatlab.org/nlab/, math.stackexchange, or even https://en.wikipedia.org/wiki/Covering_space.)

Like some other texts, even the usually clear (late) Pertti Lounesto wrote, in his “Clifford Algebras and Spinors”: “The fact that two opposite elements of the spin group Spin(2) represent the same rotation in SO(2) is expressed by saying that Spin(2) is a two-fold cover of SO(2), and written as Spin(2)/{±1} is isomorphic to SO(2).” … without giving any example.

1. In what seems like a trivial sense, the unit complex numbers exp(i(θ+2nπ)) for integers n>±1 appear to provide not just a two-fold but rather an n-fold cover of R(θ); but the relevant points in U(1) would then be identical for a given R(θ), which to me looks like 1:1 rather than 2:1 (?)

2. Another account, if I read it correctly, appears to suggest that a rotation R(θ) maps to the two points exp(±iθ) in U(1) – but that results in a reflection in the real plane z = (cos(θ) + isin(θ)) and z = (cos(θ) - isin(θ)), which doesn’t seem to agree with what someone wrote elsewhere, that "one rotation in SO(2) maps to two rotations in U(1)".

3. There’s also the case where an anticlockwise rotation by θ in the 2D plane reaches the same point as a clockwise rotation (i.e. in the alternative direction) by (θ-2π), and which yields the same final position, as (1.) above, for an anticlockwise rotation by (θ + 2π). And this is just a special case of how the same point is reached by π rotations in opposite directions.

The answer I seek is a clear and unambiguous example of the U(1) double cover of SO(2), identifying the angle θ of a single rotation R(θ) and the corresponding two (distinct?) resulting points exp(i….) and a different exp(I,,,,).

Clarification or other advice will be much appreciated.

Thank you for reading the whole question.
 
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  • #2
I think I've resolved the question and will post the proof in a day or so.
 
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  • #3
pellis said:
I think I've resolved the question and will post the proof in a day or so.
Tabulated data towards a solution for the question, based on Cl2 Clifford algebra approach that uses the factorised rotor exp() in a generalisable "sandwich product" form exp(-iθ/2)rexp(+iθ/2) to rotate a vector.

More detailed explanation to follow, once I'm satisfied that the table leads to a correct illustration of the double cover.
 

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