# $M \times N$ orientable iff both $M$ and $N$ are orientable proof in terms of volume forms [duplicate]

I'm studying differential forms, and in my homework I'm asked to show that the product of two manifolds $M \times N$ is orientable if and only if both $M$ and $N$ are orientable.

I want to show this using volume forms. For the backwards implication ($\Leftarrow$):

Suppose $M$, m-manifold, and $N$, n-manifold, are orientable. Then there exist nowhere vanishing top forms $\omega_1 \in \Omega^m(M)$ and $\omega_2 \in \Omega^n(N)$. Define $\omega_1 \times \omega_2$ in $M \times N$ by $\omega_1 \times \omega_2(X_1,...,X_m,Y_1,...,Y_n)=\omega_1(X_1,...,X_m)\omega_2(Y_1,...,Y_n)$, and one can easily see that this is a form, a top, nowhere vanishing $(m+n)$-form in $M\times N$. It follows that $M\times N$ is orientable.

My problem is in the forward implication. How do I show that if $M\times N$ is orientable, then so are $M$ and $N$, or, equivalently, if $M$ and $N$ are both not orientable, then $M\times N$ can't be orientable? How can I construct volume forms on $M$ and $N$ from a given volume form in $M\times N$?

## marked as duplicate by Brahadeesh, user10354138, Rebellos, Cesareo, KReiserDec 15 '18 at 3:25

• Your claim for the backwards implication is false. $\omega_1 \times \omega_2$ as you have defined it is not a differential form (it fails to be anti-symmetric). What you want to do is take $\pi^{*} \omega_1 \wedge \tau^{*} \omega_2$ where $\pi : M \times N \to M$ and $\tau : M \times N \to N$ are the projections. – Pedro Nov 15 '15 at 21:10
• (observe that, in your definition, $\omega_1 \times \omega_2 = \pi^{*}\omega_1 \otimes \tau^{*} \omega_2$) – Pedro Nov 15 '15 at 21:14
• The statement that you claim is equivalent to the forward implication is not equivalent. A statement equivalent to the forward implication is "if either $M$ or $N$ is not orientable, then $M\times N$ can't be orientable." – Jack Lee Nov 15 '15 at 21:22
• @Pedro Yes, after writing the question, I realized that I was taking the tensor product of the forms in my definition of $\omega_1 \times \omega_2$, and that I should take instead the wedge product. But I was missing that part about the projections. Thank you! – Strider Nov 16 '15 at 3:05
Assume $M \times N$ is orientable. Fix a point $p \in N$ and a basis $\{v_1, \ldots, v_n\}$ of $T_p N$. Consider an orientation form $\omega$ of $M \times N$ and identify $M$ with $M \times \{p\} \subset M \times N$. Now define $\eta$ on $M$ by $\eta(e_1, \ldots, e_m) = \omega(e_1, \ldots, e_m, v_1, \ldots, v_n)$.