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Can anyone show me a simple example of a manifold such that $H_{1}(\mathcal{X})=0$ but $\mathcal{X}$ is not orientable? We know the contention hold for $\pi_{1}$, I am not sure if it holds for $H_{1}$.

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In what sense do you mean orientable? You could ask for manifolds... –  Aaron Mazel-Gee Nov 4 '12 at 19:30
    
I guess I should ask for manifolds. –  Bombyx mori Nov 4 '12 at 19:32
    
If you're talking about manifolds and orientation in the usual sense, then the first thing to look for is a group $G$ to play the role of $\pi_1(X)$ whose abelianization is trivial (i.e., $[G,G]=G$) and such that $G$ contains a 2-torsion element. (Or maybe any $n$-torsion? I forget, it's been a long time since I've read Hatcher, and I'm not big into lens spaces and things...) –  Aaron Mazel-Gee Nov 4 '12 at 19:35
    
This is enough; I shall try to find one myself. Thanks. –  Bombyx mori Nov 4 '12 at 19:40
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1 Answer 1

up vote 4 down vote accepted

There are no such manifolds.

Obstruction to orientability is given by first Stiefel-Whitney class $w_1\in H^1(X;\mathbb Z/2\mathbb Z)$. So if $H^1(X;\mathbb Z/2\mathbb Z)=0$ manifold $X$ is always orientable.

In other words. If loop $\gamma$ is trivial as an element of $H_1$ it is always orientation-preserving.

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Thanks. This is clear. –  Bombyx mori Nov 4 '12 at 20:21
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