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I'm studying orientability of manifolds currently and I'm having trouble to prove the following: $M\times N$ is orientable iff $M$ and $N$ are orientable.

I am able to prove that the product is orientable if components are orientable (chart is $\{(U_\alpha\times V_{\beta},\phi_\alpha\times \psi_\beta):(\alpha,\beta)\in A\times B \}$, and $\det J=\det J_1 \det J_2>0$ by Cauchy-Binet's theorem), but I don't know how to prove the other direction.

So why this holds: if $M\times N$ is orientable, then $M$ and $N$ are orientable?

Thanks in advance.

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  • $\begingroup$ This is easy if you know some algebraic topology, like cohomology with compact coefficients and Kunneth formula. Do you know this material? $\endgroup$ Nov 3, 2013 at 19:10
  • $\begingroup$ No, I don't know. Is there some other way? $\endgroup$
    – alans
    Nov 3, 2013 at 22:16
  • $\begingroup$ math.stackexchange.com/a/1055522/3217 $\endgroup$ Mar 21, 2015 at 11:04
  • $\begingroup$ one other way also you can do by proving the existence of a non-vanishing volume form, actually orientation and existence of non-vanishing volume form is iff condition. For details you can have a look on Smooth Manifold by John Lee $\endgroup$ Nov 2, 2015 at 12:51
  • $\begingroup$ Which is the idea using compact support cohomology? $\endgroup$ Sep 22, 2022 at 13:41

2 Answers 2

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If $M\times N$ is orientable, any open submanifold is orientable. We can pick an open subset $U\subset N$ diffeomorphic to $\mathbb R^n$, and $M\times U\equiv M\times\mathbb R^n$ is orientable. By induction it is enough to see that if $M\times\mathbb R$ is orientable, then $M$ is orientable. Pick any open cover $\{W_i\}$ of $M$ such that there are diffeomorphisms $\varphi_i:\mathbb R^m\to W_i$. The cover ${\mathcal A}=\{W_i\times\mathbb R\}$ is an atlas with parametrizations $\psi_i=\varphi_i\times Id:\mathbb R^{m+1}\to W_i\times\mathbb R$. Then, if needed we can modify each $\psi_i$ by changing the sign of the first variable in $\mathbb R^{m+1}$ to make it compatible with a fixed orientation in $M\times\mathbb R$. This changes correspondingly the $\varphi_i$. Thus ${\mathcal A}$ is positive and we have $$ J(\psi_j^{-1}\circ\psi_i)=\begin{pmatrix} J(\varphi_j^{-1}\circ\varphi_i)&0\\0&1 \end{pmatrix}, $$ hence $J(\varphi_j^{-1}\circ\varphi_i)=\det J(\psi_j^{-1}\circ\psi_i)>0$. Thus the $\varphi_i$'s are a positive atlas of $M$. We are done.

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  • $\begingroup$ Could you maybe explain the "a little computation shows that $\{W_i\}$ with each $\phi_i$ correspondingly changed is a positive atlas of $M$" part a bit? What exactly is $\phi_i$ in your proof and what and how would you be computing? $\endgroup$
    – Ryker
    Nov 1, 2015 at 4:42
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    $\begingroup$ I've edited a bit, trying to make it clearer! $\endgroup$
    – Jesus RS
    Nov 2, 2015 at 13:19
  • $\begingroup$ After seeing your original answer, I figured this would be the way to go about it, but thanks for the edit and the confirmation I was on the right track. $\endgroup$
    – Ryker
    Nov 2, 2015 at 17:57
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Take an atlas $K$ of $M\times N$. Then there are parametrizations of $K$ in the form $\phi\times\psi$ where $\phi$ is a parametrization of $M$ and $\psi$ of $N$.

Statement: $A=\{\phi \text{ parametrisation of }M\text{ such that }\phi\times\psi\in K\}$ is a coherent atlas of $M$.

Indeed, if $\phi,\xi\in A$ then $\phi\times\psi,\xi\times\psi\in K$. Therefore, $\operatorname{det} J\left((\phi\times\psi)^{-1}\circ(\xi\times\psi)\right)>0$.

But, $(\phi\times\psi)^{-1}\circ(\xi\times\psi)=(\phi^{-1}\circ\xi,\operatorname{id})$.

Then the jacobian is also positive for $\phi^{-1}\circ\xi$.

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  • $\begingroup$ This answer si unclear, saying that the two product charts belong to $K$ is clearly equivalent to the thesis but you didn't prove it! $\endgroup$ Apr 25, 2022 at 15:43

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