Collecting proofs $SO(3)\cong\Bbb R P^3$

Question

There are a few questions on SE about the well-known homeomorphism $SO(3)\cong \Bbb R P^3$, but I thought it would be fun to collect as many proofs as we can find in one spot.

Background:

• $SO(3)$ is the group of orientation-preserving orthogonal transformations of $\Bbb R^3$, or equivalently the group of $3\times 3$ real matrices $A$ with $AA^\top=I$ and $\det A=1$
• $\Bbb R P^3$ is the space of lines going through the origin in $\Bbb R^4$ topologized equivalently as
  • $\left(\Bbb R^4\setminus\{0\}\right)/\sim$, where $x\sim cx$ for all $x\in\Bbb R^4\setminus\{0\}$ and $c\in\Bbb R\setminus\{0\}$ or
  • $D^3/\sim$, where $p\sim -p$ for all $p\in S^2$
• Both spaces are oriented $3$-manifolds with smooth structures
• $SO(3)$ is diffeomorphic (and thus homeomorphic) to $\Bbb R P^3$
• As varieties, $SO(3)$ is affine while $\Bbb R P^3$ is projective. Thus, they are not isomorphic as varieties.

You can find some proofs of the homeomorphism $SO(3)\cong \Bbb R P^3$ here, here, and in Hatcher $\S3.D$ (currently at the bottom of page $293$).

I will start with Hatcher's proof along with a version I've written using quaternions.

One direction I'm curious about is to think of $SO(3)$ as the Stiefel manifold $V_2(\Bbb R^3)$, and to somehow turn a pair of orthonormal vectors in $\Bbb R^3$ into a line in $\Bbb R^4$.

Answer 1

Hatcher's proof:

We define a map $\varphi\colon D^3\to SO(3)$ which takes a vector $p\in D^3$ to the rotation about the line spanned by $p$ by $|p|\pi$ radians (see $\ast$ below). If $p\in S^2$, then $\varphi(p)=\varphi(-p)$, so we have an induced $\tilde\varphi\colon \Bbb RP^3\to SO(3)$. It's clear $\tilde\varphi$ is injective since we can recover the axis of rotation and the rotation angle. Further, since every matrix $A\in SO(3)$ has either one or three real eigenvectors, every transformation must be rotation about some axis. It follows that $\tilde\varphi$ is surjective. Since $SO(3)$ is compact, $\tilde\varphi$ is a homeomorphism.

$\ast$: We use the right-hand-rule orientation convention here, so if your thumb points along $p$, your fingers wrap around in the rotation direction. This is akin to the direction of a magnetic field around a cable with an electric current.

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Quaternionic proof:

Identify $\Bbb R^4$ with $\Bbb H$ via $(a,b,c,d)\mapsto a+bi+cj+dk$.

Given a line $\ell\in\Bbb RP^3$, pick a unit quaternion $h$ which spans $\ell$ over $\Bbb R$. We define $\psi_\ell\colon \Bbb R^4\to\Bbb R^4$ by $x\mapsto hxh^*$ (where $h^*$ is the quaternionic conjugate of $h$). Since the only other choice of unit quaternion is $-h$, $\ell\mapsto \psi_\ell$ is well-defined.

Now $\psi_\ell(1)=h1h^*=|h|^2=1$, so in the basis $1,i,j,k$, the matrix of $\psi_\ell$ is $$\begin{pmatrix}1 & 0\\0 & A \end{pmatrix}.$$

On the other hand, the standard inclusion $Sp(1)\to GL(4;\Bbb R)$ lands in $SO(4)$, and this shows that the $3\times 3$ matrix $A$ is in $SO(3)$. We hence obtain $\tilde\psi\colon \Bbb RP^3\to SO(3)$. Some more work is required to show $\tilde\psi$ is a bijection, but this is at least a familiar fact to anyone familiar with computer graphics, for example.

Answer 2

Here is a completely non-explicit proof of a homeomorphism. We construct a homology equivalence $SO(3) \rightarrow \mathbb{R}P^3$, strengthen this to a homotopy equivalence, strengthen this to a simple homotopy equivalence, and strengthen this to a homeomorphism.

By calculating the cohomology of $SO(3)$ we can conclude there is a map $SO(3) \rightarrow \mathbb{R}P^\infty$ pulling back the generator of $H^1 (\mathbb{R}P^\infty ; \mathbb{Z}/2)$ to the generator of $H^1(SO(3); \mathbb{Z}/2)$. This necessarily factors through $\mathbb{R}P^3$ by cellular approximation. This new map has the property that it pulls back the generator of $H^3 (\mathbb{R}P^3 ; \mathbb{Z})$ to an odd multiple of the generator of $H^3 (SO(3) ; \mathbb{Z})$ since we can see that on $\mathbb{Z}/2$ cohomology, this pulls a generator back to a generator (seen by the ring structure).

$\mathbb{R}P^3$ has a group structure since it is a group quotient of $S^3$, we can now construct a degree 1 map by virtue of the fact that we have a degree 2 map $SO(3) \rightarrow \mathbb{R}P^3$ given by taking the degree 1 map to $S^3$ and postcomposing with the quotient to $\mathbb{R}P^3$. Simply use the group structure to take an appropriate combination of this degree 2 map and the odd degree map from before to get a degree 1 map.

By Poincare duality and the computation of the cohomologies of each space, this is then a homology equivalence.

It is easy to see that this induces an isomorphisms on fundamental groups from how the map is constructed (check it on first $\mathbb{Z}/2$ cohomology and use the fact that first cohomology is dual to the fundamental group). Since both of these spaces are simple (as they are groups), we deduce this homology equivalence is actually a homotopy equivalence.

In fact this homotopy equivalence must be a simple homotopy equivalence, since by calculation the Whitehead group for $\mathbb{Z}/2$ is trivial.

Finally, a simple homotopy equivalence of 3 manifolds is always homotopic to a homeomorphism by work of Turaev: "Towards the topological classification of geometric 3-manifolds".