Orientability of $\mathbb{RP}^3$

I was wondering if there is a nice way to see that $\mathbb{RP}^{3}$ is orientable without using tools of algebraic topology, like homology.

The only think I could think of was to argue that $\mathbb{RP}^{3}=\mathbb{R}^3 \cup \mathbb{RP}^{2}$ and perhaps you could argue that to get back to any starting position you have to cross the $\mathbb{RP}^{2}$ boundary but I'm pretty sure that what I'm thinking is nonsense.

This was a question on the homework for one of my topics courses and I plan on asking the professor about it tomorrow, but I was curious to see if anyone had any interesting ways of thinking about or picturing this space.

Solution 1:

The orientation on the universal cover, $S^3$, descends. It takes some checking of details, but you can do it for $\mathbb{R}P^n$ in exactly the same way for $n$ odd. Actually, much more generally if you have a covering $\pi: X\to B$, and $Aut_\pi (X)$ is cyclic and generated by some orientation preserving diffeo $f: X\to X$, then if $X$ is orientable, the orientation will descend to $B$. In this case the diffeo is the antipodal map which is orientation preserving on $S^n$ if and only if $n$ is odd.

Solution 2:

One can also see that $\mathbb{R}P^3$ is a Lie group, and that all Lie groups are orientable.

To see that $\mathbb{R}P^3$ is a Lie group, first think of $S^3$ as the collection of unit quaternions. The subset $\{1,-1\}$ is not only a subgroup, but it is a normal subgroup of $S^3$. Hence, $S^3/\{1,-1\}$ is a Lie group. In fact, the elmement -1 acts as the antipodal map. (It turns out, that as a Lie group, $\mathbb{R}P^3$ is isomorphic to $SO(3)$, the 3x3 real matrices satisfying $AA^t = A^tA = Id$.)

Why is every (connected component of a) Lie group orientable? Well, suppose $G$ is a (connected) Lie group with identity $e$. At the identity, pick any ordered basis $e_i$ for $T_e G$ (i.e., pick an orientation on the vector space $T_e G$. To figure out what what orientation at $g\in G$ should be, notice the map $L_g:G\rightarrow G$ given by $L_g(h) = gh$ maps $e$ to $g$. Hence its differential $d_e L_g$ maps $T_e G$ to $T_g G$.

The map $L_g$ is actually a diffeomorphism (with inverse $L_{g^{-1}}$), so the map $d_e L_g$ must also be an isomorphism. Now, simply take as an ordered basis $d_e L_g e_i$.

Solution 3:

$RP3$ can be represented as a three dimensional ball $D3$ with the antipodal points of the bounding sphere identified. One can draw a small loop on the surface of the bounding sphere and compute the direction of the normal according to the right hand rule. Now, performing the same computation for the antipodal loop, the normal will have the same direction. Thus there is a consistent choice of the normal. Notice that the same argument will fail in the case of the nonorientable $RP^2$.