Integration over non-orientable Riemannian manifold I am confused a bit by the following theorem that integrates over non-orientable manifold. How is that possible? (see this post).
Theorem. Let $M^n$ be a compact Riemannian non-orientable manifold, and $\omega$ a
differential 1-form. Let $\pi:(\widetilde{M},\tilde{g})\to (M,g)$ be the two-sheeted covering map, where $\tilde{g}=\pi^*g$ and let $\widetilde{A}=\pi^*A$. Then
$$\int_\widetilde{M}\tilde{\delta} \widetilde{\omega}\widetilde{dV}=2\int_M\delta\omega dV.$$
 A: It suffices to consider the case when $M$ is connected. Since all what you care about are integrals, just choose an open topological disk $U\subset M$ whose complement has measure zero, then pick an orientation on $U$ any way you want. For instance, you can take $U$ to be the complement to the cut-locus of a point in $M$. (If you do not like this, observe that $M$ is a connected sum of an oriented surface and the projective plane. Hence, $M$ contains a 1-sided smooth loop $c$ such that $M'=M-c$ is orientable. You can use $M'$ instead of a disk $U$.)
This will define a volume form $dV$, the Hodge star and $\delta$ on $U$. Then check that the integrand
$$
\delta\omega dV
$$
is independent of the chosen orientation on $U$ (you will be changing the sign twice) and the integral
$$
\int_U \delta\omega dV 
$$
is independent of the choice of $U$ (since $M\setminus U$ has measure zero). This is how you make sense of    the integral
$$
\int_M \delta\omega dV.  
$$
Now, consider the 2-fold covering $\pi$ and observe that $\pi^{-1}(U)$ has two components $U_1, U_2$ such that $\pi|_{U_i}: U_i\to U$ is a diffeomorphism, $i=1,2$. Both integrals
$$
\int_{U_i} \tilde\delta\tilde\omega \widetilde{dV}$$
will equal to
$$
\int_U \delta\omega dV. 
$$
Thus, you get your formula.
The same trick works in higher dimensions as well.
