# Product of manifolds & orientability

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?

• This is easy if you know some algebraic topology, like cohomology with compact coefficients and Kunneth formula. Do you know this material? Nov 3, 2013 at 19:10
• No, I don't know. Is there some other way? Nov 3, 2013 at 22:16
• math.stackexchange.com/a/1055522/3217 Mar 21, 2015 at 11:04
• 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 Nov 2, 2015 at 12:51
• Which is the idea using compact support cohomology? Sep 22, 2022 at 13:41

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.

• 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? Nov 1, 2015 at 4:42
• I've edited a bit, trying to make it clearer! Nov 2, 2015 at 13:19
• 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. Nov 2, 2015 at 17:57

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 $$\phi^{-1}\circ\xi$$.

• 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! Apr 25, 2022 at 15:43