# 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.

• Have you tried using the usual atlas and checking if the determinants of the Jacobians of the transition functions are positive? Oct 5, 2010 at 13:01
• Our professor said he didn't want us to think of the differential geometry style definition (which I neglected to mention in my question) and instead to argue that it suffices to check the antipodal map is orientation preserving and that we'd talk more about it next class. I've also only seen atlases and manifolds from informal and intuitive approaches and had a hard time seeing why the determinant of the Jacobian couldn't be negative. Oct 5, 2010 at 21:05
• You just have to compute the determinants of the Jacobians... that is not "hard to see" :) Oct 6, 2010 at 15:39

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.

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 $d_e L_g e_i$.

• Out of curiosity: How can we prove that this gives a well-defined orientation on $G$ in the sense that it is induced by the charts of an atlas? Naively, I'm thinking we can just take any chart $\sigma\colon U\subseteq \mathbb{R}^m\to G$ with $e\in G$, and use left-translations to get an atlas $\{L_g\circ\sigma:g\in G\}$. But does this give a well-defined orientation? In other words, if $h\in L_g(U)$ for some $g\in G$, does the charts $L_g\circ\sigma$ and $L_h\circ\sigma$ induce the same orientation on $T_hG$, i.e. does it holds that $\det(\sigma^{-1}\circ L_{g^{-1}h}\circ\sigma)>0$? Aug 14, 2020 at 15:44
• Once you have a choice of ordered basis at any point (i.e., once you know a manifold $M$ is parallellizable), you can define orientation-via-charts by declaring $(U,\phi)$ to be "correctly" oriented if the orientation $(\frac{\partial}{\partial x_1},... ,\frac{\partial}{\partial x_n})$ "agrees" with the preassigned ordered basis, in the sense that the transition matrix from the basis of partials to the preassigned basis has positive determinant. If $(U,\phi)$ is incorrectly oriented, one can precompose with an orientation reversing map $\mathbb{R}^n\rightarrow \mathbb{R}^n$ to make it corect. Aug 14, 2020 at 16:17

$RP3$ 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 $RP^2$.

$RP^{2k-1}$ is a quotient of a codimension 1 submanifold with induced orientation ($S^{2k-1}$)in $R^{2k}$, by an orientation-preserving transformation of $R^{2k}$. Because there is a consistent notion of normal vector (i.e., of "inside" and "outside") on the submanifold, preserving orientation of the surrounding manifold also preserves orientation on the submanifold. Transporting the orientation along a path between two points of the submanifold that are identified in the quotient can be viewed as transporting the orientation in the surrounding manifold, and this has to be consistent (have positive determinant on the local frames) because of the orientability of the surrounding manifold.