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I was reading a paper and it was affirmed in there that $\mathbb{R}\mathbb{P}^2\times\mathbb{S}^1$ was a non-orientable 3-manifold.

Does anyone knows how to prove it? if not, is there another (simple) example of a non-orientable 3-manifold?

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  • $\begingroup$ Do you believe that $\mathbb{R}P^2$ is non-orientable? $\endgroup$ – Randall Mar 15 '18 at 16:00
  • $\begingroup$ yes (actually I saw the proof that $\mathbb{R}\mathbb{P}^n$ is non-orientable if $n$ is even) @Randall $\endgroup$ – Pires Dankan Mar 15 '18 at 16:03
  • $\begingroup$ If you can take $H_1$ you are now done. $\endgroup$ – Randall Mar 15 '18 at 16:06
  • $\begingroup$ Is there any other way to prove it without using Homology group? @Randall $\endgroup$ – Pires Dankan Mar 15 '18 at 16:11
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It has a connected orientable double cover, which is $S^2\times S^1$. Thus the space itself must be non-orientable.

Or you could use the fact that the product of manifolds $X$ and $Y$ is orientable if and only if both $X$ and $Y$ are orientable.

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  • $\begingroup$ $S^2 \times S^1$ also has a connected orientable double cover (namely $S^2 \times S^1$)... as $\Bbb Z$ has a subgroup of index 2. $\endgroup$ – PVAL-inactive Mar 16 '18 at 8:44
  • $\begingroup$ @PVAL-inactive that is true. But it is not an orientation covering. The orientation covering of $S^2 \times S^1$ is its disjoint union with itself. $\endgroup$ – Mihail Mar 16 '18 at 11:29

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