Timeline for Why do I need densities in order to integrate on a non-orientable manifold?
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| Oct 24, 2023 at 17:06 | comment | added | lightxbulb | Would all the integrals of differential forms that rely on the orientation be zero on the double cover? E.g. integrating flux over a Mobius strip embedded in $\mathbb{R}^3$: $\omega = \star(f_1 dx + f_2 dy + f_3 dz)$. Then if I use the standard parametrization for a Mobius strip I would change $\theta\in [0,2\pi]$ to $\theta\in[0,4\pi]$. But then at every point I will once pass through a positive normal, and once through a negative one, which would cancel out and give me zero. So is this mostly useful for computing integrals not relying on orientation? | |
| Nov 13, 2020 at 22:08 | comment | added | C.F.G | Actually this integral comes from the theory of the Lebesgue integral which is based on the measure. | |
| Aug 16, 2017 at 15:00 | history | answered | Qfwfq | CC BY-SA 3.0 |