I'm learning differential geometry for the first time. I'm sure this question is answered in a number of books, but it doesn't seem to be addressed in any book I have.
Let $M,N$ be orientable manifolds of dimension $m,n$. I'd like to view an orientation on $M$ as given by a choice of a nowhere vanishing top differential form. The product maps $$N\leftarrow M\times N\rightarrow M$$ give maps $\Omega^n(N)\rightarrow \Omega^n(M\times N)$ and $\Omega^m(M)\rightarrow\Omega^m(M\times N)$.
Thus given an orientation $n$-form $\omega_N$ on $N$ and an orientation $m$-form $\omega_M$ on $M$, we get two natural choices of orientation forms on $M\times N$, namely $\omega_N\wedge \omega_M$ and $\omega_M\wedge\omega_N$. They are the same if and only if $M,N$ are both even-dimensional. (Here I don't want to distinguish $M\times N$ from $N\times M$).
In particular, if either $M$ or $N$ are odd-dimensional, then this doesn't seem to give a canonical choice of an orientation on the product $M\times N$.
Would it be correct to say that the "product orientation" on the product of two oriented manifolds only exists if both manifolds are even dimensional? Ie, if one of the manifolds is odd-dimensional, then there is no canonical choice of an orientation on the product?