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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!

Hang
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2 Answers2

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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.

Lukas Geyer
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    An alternative, if you don't know the fixed point theorem, is the following: Every map of the sphere $S^n$ without fix points is homotopic to the antipodal map, hence has degree $(-1)^{n+1}$. Therefore, any degree $-1$ self map of an odd-dimensional sphere must have a fixed point. – archipelago Jun 02 '15 at 13:47
  • @Lukas Geyer :Thank you! But I am a bit lost. Do you mean " degree of +1 $\Rightarrow$ preserve orientation "? And could you please explain why this implies that X is orientable? – Hang Jun 02 '15 at 14:43
  • @Henry: Deck transformations are homeomorphisms, with degree $\pm 1$ depending on whether they preserve or reverse orientation, so in this particular case degree 1 is equivalent to preserving orientation. – Lukas Geyer Jun 02 '15 at 15:08
  • @Henry For your second question see http://math.stackexchange.com/questions/163502/m-gamma-is-orientable-iff-the-elements-of-gamma-are-orientation-preserving – archipelago Jun 02 '15 at 15:14
  • Your referenced question works for smooth manifold. But I guess here $X$ should be just a topological manifold. – Hang Jun 03 '15 at 11:23
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    @Henry: Unless I am missing something, I don't see how this argument uses smoothness at all. Where do you see a problem? – Lukas Geyer Jun 03 '15 at 15:50
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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.

Alex Fok
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  • I'm a bit lost as to where you're getting the formula $H^\ast (X, \mathbb{R}) = H^\ast(S^{2n+1},\mathbb{R})^\Gamma$ from. Can you explain that a bit more, or give a reference? – Robert Short Jun 02 '15 at 13:48
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    i suppose you are talking about the formula $H^(X, \mathbb{R})=H^(S^{2n+1}, \mathbb{R})^{\Gamma}$. This can be proved by looking at the cochain complexes defining the cohomology and using a simple averaging argument (note that this averaging is only possible for a finite group). – Alex Fok Jun 02 '15 at 13:53
  • Could you please tell me the meaning of your notation $H^∗(S^{2n+1},\mathbb{R})^Γ$? I am not familiar with this. – Hang Jun 02 '15 at 14:10
  • It means the subspace of $H^(S^{2n+1}, \mathbb{R})$ consisting of those elements $\alpha$ such that $\gamma^\alpha=\alpha$ for all $\gamma\in \Gamma$. – Alex Fok Jun 02 '15 at 14:13
  • @RobertShort A reference I can think of is Bott and Tu, Differential forms in algebraic topology, p.77, exercise (a). Though it deals with a special case where the group is of order two, and is in the language of de Rham cohomology, it can be generalized to any finite group and adapted to the singular cohomology case easily. Also the hint in part (b), which is the averaging argument I mentioned, is useful in proving the infectivity of the isomorphism. You also need the fact that any invariant form on the covering space can be pushed down to a form on the quotient space in order to show onto. – Alex Fok Jun 02 '15 at 14:24
  • Very nice, thank you @AlexFok ! – Robert Short Jun 03 '15 at 11:18
  • Perhaps your answer were too advanced for me. I try hard but still cannot understand. For example, why the Euler characteristic of $X$ should be that? How your formula$H^(X, \mathbb R)= H^ (S^{2n+1},R)^Γ$ comes? and finally how $\chi(X)=0$ implies $H^{2n+1}= \mathbb R$? Sorry for so many question, could you please give more details? – Hang Jun 03 '15 at 11:38