Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 2 down vote favorite

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 $H_{1}$.

share|improve this question
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. – user32240 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. – user32240 Nov 4 '12 at 19:40

1 Answer

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.

share|improve this answer
Thanks. This is clear. – user32240 Nov 4 '12 at 20:21

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.