# If $n$ is odd then $\mathbb{R}P^n$ is orientable

Show that if $n$ is odd then $\mathbb{R}P^n$ is orientable.

Comments: I have the following: The antipodal map $\alpha: S^n \longrightarrow S^n$, $\alpha (x) = -x$ is orientation-preserving if and only if n is odd. I'm trying to use the map $\pi: S^n \longrightarrow \mathbb{R}P^n$ and the above fact to find orientation form of $\mathbb{R}P^n$. But I can not fit these facts. What is the orientation form?

• Use the induced map $\pi^* :\Omega^n(\mathbb{R}P^n)\to \Omega^n(S^n)$ : what's its image ? Jul 4 '18 at 21:00
• If $n$ odd, what will be a orientation form of $\mathbb{R}P^n$ ? Jul 5 '18 at 2:27
• @I4teLearner : no, that's not quite right : any pulled back form is invariant under the action of the antipodal map, so any form that isn't invariant isn't pulled back. The point of the suggestion is to show that this is in fact an equivalence May 11 at 7:22
• @I4teLearner : this is essentially Lee Mosher's answer. Note that if $\pi^*\omega$ is nowhere vanishing and $\pi$ is surjective, then $\omega$ is also nowhere vanishing. So by the fact that the image contains a nonvanishing form, there must also be a nonvanishing top form on $\mathbb RP^n$ May 11 at 18:14
• @I4teLearner: oh I didn't realize that was your question - I thought you knew how to do that from your earlier comment. For that you can either look at Lee Mosher's comment below the answer here; or you can look at my answer there : math.stackexchange.com/questions/2840494/… - it's a slightly different case but exactly the same proof May 12 at 9:08

You seem most interested in the case where $$n$$ is odd, and proving in that case that $$\mathbb RP^n$$ is orientable, so I'll provide an answer for that case.
Fix an orientation on $$S^n$$, which means that for each $$x \in S^n$$ we have an orientation $$\mathcal O_x$$ of the tangent space $$T_x S^n$$, and these orientations vary continuously as $$x \in S^n$$ varies.
To define an orientation on $$\mathbb RP^n$$, consider an arbitrary point of $$\mathbb RP^n$$, which I'll represent as unordered pair of antipodal points of $$S^n$$. Using equivalence class notation, for each $$x \in S^n$$ I'll write this as $$[x]=[-x]=\{x,-x\}$$. So, we have to define an orientation $$\mathcal O_{[x]}$$ of the tangent space $$T_{[x]} (\mathbb RP^n)$$: define $$\mathcal O_{[x]} = D_x\pi(\mathcal O_x)$$ The key issue is whether this is well-defined independent of the choice of the two representatives $$x$$ and $$-x$$ of $$[x]$$, and that is true because \begin{align*} D_{-x}\pi \, (\mathcal O_{-x}) &= D_{-x}(\pi \circ \alpha) \, (\mathcal O_{-x}) \\ &= D_{\alpha(-x)} \pi \circ D_{-x} \alpha \, (\mathcal O_{-x}) \\ &= D_x \pi \circ D_{-x} \alpha (\mathcal O_{-x}) \\ &= D_x \pi \, (\mathcal O_{x}) \end{align*} where the first equation holds because $$\pi = \pi \circ \alpha$$, the second equation is the chain rule, the third equation uses that $$\alpha(-x)=x$$, and the fourth equation uses that $$\alpha$$ preserves orientation and so $$D_{-x}\alpha(\mathcal O_{-x}) = \pi(\mathcal O_x).$$
• @Lee Mosher. How are you sure that this orientation on the tangent space at $\mathbb{R}P^n$ varies continuously? Is this just inherited?
• The equation $\mathcal O_{[x]} = D_x\pi(\mathcal O_x)$ literally says that the orientation on $\mathbb RP^n$ at the point $[x]$ is inherited, via the derivative map $D_x$, from the given orientation $\mathcal O$ on $S^n$ at the point $x$. Continuity of the orientation $\mathcal O_{[x]}$ on $\mathbb RP^n$ then follows from continuity of the orientation $\mathbb O_x$ on $S^n$ (after well-definedness has been checked), by applying that formula in any compatible coordinate charts for $S^n$ and $\mathbb RP^n$. Jan 5 '20 at 15:37