9
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ This is easy if you know some algebraic topology, like cohomology with compact coefficients and Kunneth formula. Do you know this material? $\endgroup$ – Moishe Kohan Nov 3 '13 at 19:10
  • $\begingroup$ No, I don't know. Is there some other way? $\endgroup$ – alans Nov 3 '13 at 22:16
  • $\begingroup$ math.stackexchange.com/a/1055522/3217 $\endgroup$ – Georges Elencwajg Mar 21 '15 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 $\endgroup$ – Anubhav Mukherjee Nov 2 '15 at 12:51
13
$\begingroup$

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 $\psi_i=\varphi_i\times Id:\mathbb R^{n+1}\to W_i\times\mathbb R$. Then, if needed we can modify each $\psi_i$ by changing the sign of the first variable in $\mathbb R^{n+1}$ 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 $$ J(\psi_j^{-1}\circ\psi_i)=\begin{pmatrix} J(\varphi_j^{-1}\circ\varphi_i)&0\\0&I \end{pmatrix}, $$ 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.

$\endgroup$
  • $\begingroup$ 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 '15 at 4:42
  • 1
    $\begingroup$ I've edited a bit, trying to make it clearer! $\endgroup$ – Jesus RS Nov 2 '15 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 '15 at 17:57
0
$\begingroup$

Take an atlas K of MxN. Then the parametrizations of K be in the form $\phi\times\psi$ where $\phi$ is a parametrization of M and $\psi$ of N.

Statement: A={$\phi$ param of M such that $\phi\times\psi\in K$} is an atlas coerent of M.

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

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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