note-math

Euclidean $ℝ^{2}$ 's direction space is $𝕊 = 𝕊^{1}$

Rotation is the part of $ℝ^{2}$ 's (direction-preserving) #link(<isometry>)[] that preserves the direction space $𝕊$

The isometry of $ℝ^{2}$ is $SO (2) ⋊ ℝ^{2}$ (it can be proven that $ℝ^{2}$ isometry implies #link(<affine>)[affine])

Rotation is $SO (2)$

The element of $SO (2)$ $(\begin{matrix} 𝑎 & - 𝑏 \\ 𝑏 & 𝑎 \end{matrix})$ with $𝑎^{2} + 𝑏^{2} = 1$ . Set-theoretically equivalent to $𝕊$

Also compatible in multiplication $(\begin{matrix} 𝑎 & - 𝑏 \\ 𝑏 & 𝑎 \end{matrix}) (\begin{matrix} 1 \\ 0 \end{matrix}) = (\begin{matrix} 𝑎 \\ 𝑏 \end{matrix}) ≃ (\begin{matrix} 𝑎 & - 𝑏 \\ 𝑏 & 𝑎 \end{matrix}) = (\begin{matrix} 𝑎 & - 𝑏 \\ 𝑏 & 𝑎 \end{matrix}) (\begin{matrix} 1 \\ 1 \end{matrix})$

The multiplication of elements in $SO (2)$ is equivalent to the multiplication of elements of length $1$ in $ℂ$ . recall $ℂ$ is #link(<normed-algebra>)[] $| 𝑎 𝑏 | = | 𝑎 | | 𝑏 |$

Question angle_(tag)

Probably not the perfect motivation

Restricting the $ℝ^{2}$ metric to $𝕊$ yields #link(<metric-manifold>)[]

Intuitively, in Euclidean $ℝ^{2}$ , we can "rotate", and the composition of rotations corresponds to the addition of "angles"

The latter should be the $exp : ℝ \to Isom 𝕊$ of $𝕊$ 's #link(<Killing-field>)[] as a one-parameter homomorphism to $𝕊$ 's isometry

Calculate $exp$ with #link(<geodesic>)[geodesic]. Calculate geodesic with, for example, raw("stereographic projection coordinates"). For the geodesic starting at $(\begin{matrix} 1 \\ 0 \end{matrix})$ , the result is denoted as trigonometric-function_(tag) trigonometric function $(\begin{matrix} cos (𝑡) \\ sin (𝑡) \end{matrix})$ . The power series expansion of $(\begin{matrix} cos (𝑡) \\ sin (𝑡) \end{matrix})$ at $𝑡 = 0$ can be calculated using #link(<inverse-analytic>)[inverse function theorem]

\begin{matrix} cos (𝑡) & = & \sum \frac{{(- 1)}^{𝑛}}{(2 𝑛)!} 𝑡^{2 𝑛} \\ sin (𝑡) & = & \sum \frac{{(- 1)}^{𝑛}}{(2 𝑛 + 1)!} 𝑡^{2 𝑛 + 1} \end{matrix}

Homomorphism is reflected in, according to power series

(\begin{matrix} cos (𝑠 + 𝑡) & - sin (𝑠 + 𝑡) \\ sin (𝑠 + 𝑡) & cos (𝑠 + 𝑡) \end{matrix}) = (\begin{matrix} cos (𝑠) & - sin (𝑠) \\ sin (𝑠) & cos (𝑠) \end{matrix}) (\begin{matrix} cos (𝑡) & - sin (𝑡) \\ sin (𝑡) & cos (𝑡) \end{matrix})

Or using Euler-formula_(tag) $exp i 𝑡 = cos 𝑡 + i sin 𝑡 ≃ (\begin{matrix} cos (𝑡) & - sin (𝑡) \\ sin (𝑡) & cos (𝑡) \end{matrix})$ , then using $ℂ$ #link(<exponential>)[exponential function] and complex number multiplication $exp i (𝑠 + 𝑡) = exp i 𝑠 \cdot exp i 𝑡$

Thus $ℝ ↠ SO (2) ≃ 𝕊 ≃ U (1, ℂ)$

Hyperbolic angle is the same $ℝ \leftrightarrow SO (1, 1) ≃ ℍ 𝕪 ≃ U (1, ℂ_{split})$