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- Product of manifolds & orientability 2 answers
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 $M\times N$?