A Geometric Exploration of Space: SO(3), RP(3), and the 3-Sphere

May 5—2025

The motivation for this post comes from the example Arnold gave on p. 81 in his PDE book. My goal is to visualize the isomorphism between SO(3) and $\mathbb{R}P^3$ as well as the mapping from $S^3$ to SO(3).

The key is to find a common geometric object that is easy for us to see as isomorphic to SO(3) and $\mathbb{R}P^3$ .

We can express SO(3) as:

SO(3) = ${(\overrightarrow{v}, \alpha) : \overrightarrow{v} \in S^2, \alpha \in [0, \pi]}$ . Note: $(\overrightarrow{v}, \pi)$ and $(-\overrightarrow{v}, \pi)$ are the same and the identity map $I$ can be identified as ${(\overrightarrow{v}, 0): \overrightarrow{v} \in S^2}$ .

I will begin by defining a new geometric object, the solid ball in $\mathbb{R}^3$ with the antipodal points on its boundary identified as the same point. Let’s call this object with radius $\pi$ $\sim{B}$ .

Claim: We can find an isomorphism between $\tilde{B}$ and SO(3) and $\mathbb{R}P^3$ , respectively.

Isomorphism between SO(3) and $\tilde{B}$

We define $\phi: SO(3) \rightarrow \tilde{B}$ as the following:

$\phi((\overrightarrow{v}, \alpha)) = \overrightarrow{v} \cdot \alpha$ when $\alpha \neq 0$

and $\phi(I) = (0, 0, 0)$ .

This map is well-defined because $\overrightarrow{v} \cdot \alpha$ and $-\overrightarrow{v} \cdot \alpha$ are identified as the same just as $(\overrightarrow{v}, \pi)$ and $(-\overrightarrow{v}, \pi)$ are identified as the same. It is not difficult to see that $\phi$ is a bijection between $\tilde{B}$ and SO(3).

Isomorphism between $\mathbb{R}P^3$ and $\tilde{B}$

The map that I have found is a bijection, but am not uncertain if it shows the “isomorphism” between $\mathbb{R}P^3$ and $\tilde{B}$ (because to be honest, what structure do the two share?).

Recall, ${(x_1, x_2, x_3, x_4), (-x_1, -x_2, -x_3, -x_4)} = [(x_1, x_2, x_3, x_4)] \in \mathbb{R}P^3$ .

Define $\phi : \mathbb{R}P^3 \rightarrow \tilde{B}$ by

$\phi([(x_1, x_2, x_3, x_4)]) = (x_1 \cdot x_4, x_2 \cdot x_4, x_3 \cdot x_4)$ when $x_4 \neq 0$

and $\phi([(x_1, x_2, x_3, x_4)]) = [(x_1, x_2, x_3)]$ when $x_4 = 0$ (this is because $x_4 = 0$ , $(x_1, x_2, x_3) \in \partial \tilde{B}$ ).

It is not difficult to see that this map is a bijection.

A double-cover map from $S^3$ to $\tilde{B}$

Define $\tau : S^3 \rightarrow \tilde{B}$ by:

$\tau((x_1, x_2, x_3, x_4)) = (x_1, x_2, x_3)$ . Note: when $x_1^2 + x_2^2 + x_3^2 = \pi^2$ , $\tau((x_1, x_2, x_3, 0)) = {(x_1, x_2, x_3), (-x_1, -x_2, -x_3)} = \tau((-x_1, -x_2, -x_3, 0))$ .

When $x_1^2 + x_2^2 + x_3^2 < \pi^2$ , we have ${(x_1, x_2, x_3)}^{-1} = { (x_1, x_2, x_3, \sqrt{\pi^2 - (x_1^2 + x_2^2 + x_3^2)}), (x_1, x_2, x_3, -\sqrt{\pi^2 - (x_1^2 + x_2^2 + x_3^2)})}$ .