A space $X$ with $S^{2n+1}$ as universal covering space must be orientable.

Question

Problem If $X$ has $S^{2n+1}$ as universal covering space, then show that $X$ must be orientable.

My idea: By contradiction, suppose $X$ is non-oreintable. Then we consider the orientation covering of $X$, that is, we have a 2-sheet covering, then there must exists a subgroup $H$ of index 2.

On the other hand, it is known that the group of deck transformation, denoted by $G$, is isomorphic to $\pi_1(X)$. Hence, we try to find some information from $G$ since all elements in $G$ gives a map of $S^{2n+1}$

What's more, I don't know how to use the odd dimension $2n+1$

Thank you!

Answer 1

By the Lefschetz fixed point theorem, every map of degree $-1$ of an odd-dimensional sphere to itself has a fixed point. However, deck transformations have no fixed points, so in this case they have to have degree $+1$ and preserve orientation.

Answer 2

Suppose $\Gamma$ is the group of deck transformations. As $S^{2n+1}$ is compact, $|\Gamma|$ must be finite, and let it be $d$. We have $X=S^{2n+1}/\Gamma$. The Euler characteristic of $X$ should be $\displaystyle\frac{\chi(S^{2n+1})}{d}=0$. Using the fact that taking cohomology with real coefficient and taking invariants under a finite group commute, we have \begin{eqnarray} H^*(X, \mathbb{R})=H^*(S^{2n+1}, \mathbb{R})^{\Gamma} \end{eqnarray} So $H^0(X, \mathbb{R})=\mathbb{R}$, $H^i(X, \mathbb{R})=0$ for $0<i<2n+1$. As $\chi(X)=0$, $H^{2n+1}(X, \mathbb{R})=\mathbb{R}$. That means $X$ is orientable. So $\Gamma$ is orientation-preserving.