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?
Best Answer
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 $D_{-x}\alpha(\mathcal O_{-x}) = \pi(\mathcal O_x).$