If a smooth manifold X is covered by an odd sphere, then X is orientable.

Question

In solving some old qualifying exam questions, I've been thoroughly stumped.

If a smooth manifold $X$ is covered by an odd dimensional sphere, then $X$ is orientable.

I see this question has been posed here, but it received little attention. The one answer I did see was not particularly illuminating.

I hoped that if $X$ was not orientable, then its orientation double cover is connected, and hence by the universal property also covered by $S^{2n+1}$. I'm also aware that $\chi(X)=0$, but I'm not sure how to proceed. Regrettably, this problem is not as easy as the case where the space is covered an even sphere, since this permits you to employ the fact that $Z_2$ is the only group acting freely on $S^{2n}$ (and hence must be $RP^n$ or $S^{2n}$).

Thanks in advance for your help!

Answer 1

I assume you mean a smooth covering map. If so, recall that if $\tilde{X}$ is a connected, oriented, smooth manifold and $\pi:\tilde{X}\to X$ is a smooth normal covering map. Then $X$ is orientable if and only if every deck transformation is orientation preserving on $\tilde{X}$. (Lee, Smooth Manifolds, pg.392)

Say that $\pi:S^{2n+1}\to X$ is a smooth covering map. As $S^{2n+1}$ is simply connected, it is the universal cover of $X$ and is, in particular, a smooth normal covering map. Since $S^{2n+1}$ is connected and oriented, the theorem reduces to showing that every deck transformation is orientation preserving. Since the deck transformations act freely on $S^{2n+1}$, every non-identity deck transformation has no fixed points. Thus, in the spirit of Xipan's answer, every deck transformation is homotopic to the antipodal map, which has degree $1$. In particular, every deck transformation is orientation preserving.

Answer 2

Any non-trivial deck transformation $f$ has no fixed point. $\forall p \in S^{2n+1}$, $p \ne f(p)$. Define $F: S^{2n+1}\times[0,1] \rightarrow S^{2n+1}$ by $$F(p,t)=\frac{(1-t)f(p)-tp}{|(1-t)f(p)-tp|}$$Since $p\ne f(p)$ this is well defined and smooth. Intuitively, since $p\ne f(p)$, there is a unique path( the shortest path, which is a geodesic) from $f(p)$ to the antipodal point $-p$, $F$ is just sliding $f(p)$ to $-p$ along this path. $F$ is a smooth homotopy so $\deg{f}=\deg(-id)=(-1)^{2n}=1$, so $f$ preserves orientation.

Push down this $f$ preserving orientation we get an orientation on the base space.