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In textbooks Stokes' theorem is usually formulated for orientable manifolds (at least I couldn't find any version not using orientability). Is Stokes theorem: $\int\limits_{M}d\omega=\int\limits_{\partial M} \omega$ also true for non-orientable manifolds? Are there any references, you can tell me?

Regards.

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    $\begingroup$ Depending on what you intend to do, you could pull-back the forms to the oriented double-cover, where Stokes's theorem applies. $\endgroup$ Mar 14, 2013 at 8:47
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    $\begingroup$ My first attempt would also be the one that José suggests. A good source for this method (imho) is Allen Hatcher's book 'Algebraic Topology', p.234, which is available at Hatcher's homepage. Of course if your problem is local enough you can just restrict to yourself to a chart (which is orientable) or try to find some large orientable subdomain from your manifold. $\endgroup$ Mar 14, 2013 at 8:58
  • $\begingroup$ You need orientability to define the integral, so the form $\omega$ would have to be a twisted differential form - with values in some kind of orientation bundle so that $\omega$ has an integral over $\partial M$ and $d\omega$ (whose existence now requires a connexion on this bundle) has an integral over $M$. Presumably, if the definitions are correct, harvesting the formula will be easy. $\endgroup$
    – ACL
    Mar 14, 2013 at 9:22

2 Answers 2

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This is just more information on Jose's comment. Look at section 10 (pages 122 to 128) in:

Peter W. Michor: Topics in Differential Geometry. Graduate Studies in Mathematics, Vol. 93 American Mathematical Society, Providence, 2008. (pdf)

On the orientable double cover $\tilde M$ of $M$ (given by $\lbrace \xi\in Or(M): |\xi|=1\rbrace$ using 10.7) there are two kinds differential forms, namely the eigenspaces for $\pm 1$ under the action of the covering map. These are the "formes paire" and "formes impaire" on $M$ in the book of de Rham; for these you formulate Stokes' theorem directly on $M$, and it is the same as the Stokes' theorem on $\tilde M$. Note that you also have the corresponding double cover of the boundary.

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This is mostly an elaboration on Peter Michor's answer, but I think more deserves to be said, including a major caveat. A reference which treats this material more fully and explicitly than his book cited above is Theodore Frankel's The Geometry of Physics.

From the "double-cover" perspective, I believe the usual differential forms are the ones in the $+1$ eigenspace of the covering map, which de Rahm calls paire ("even"). The ones in the $-1$ eigenspace ("odd forms" impaire) are called "pseudo-forms" by Frankel, although Frankel doesn't explicitly work in terms of the double cover. See section 2.8.e of Frankel for his definitions and section 3.4 for a discussion of integration, including a statement of Stokes' theorem for odd forms. (I had trouble finding such a statement in Michor's book...)

The caveat I want to mention is that like the familiar even forms, odd forms do require some orientation data to integrate over a submanifold (as in Stokes' Theorem). It's just different data. Whereas an even form is integrated over an oriented submanifold, an odd form is integrated over a transverse-oriented submanifold. That is, if $N^k \subset M^n$ and $\omega$ is an even $k$-form, then you need to choose an orientation on the tangent bundle of $N$ before you can integrate $\omega$ over $N$. If $\omega$ is an odd form, then you need to choose an orientation on $N$'s normal bundle in order to integrate (a fair number of arbitrary need to be made to specify such a thing, including, at least locally, a choice of normal bundle, but they turn out not to matter).

This treatment of orientation means that, for example, an odd $(n-1)$ form on an $n$-manifold can naturally represent a "flux": integrate over a hypersurface to determine the rate of transport through that hypersurface per unit time -- the transverse orientation tells you which direction of flow you consider to be "positive".

But the case that comes out most naturally is when you have an odd $n$-form on an $n$-manifold. In this case, when you integrate over an $n$-submanifold, the normal bundle is 0-dimensional, so there is a canonical choice of transverse orientation (between "$+$" and "$-$", just take "$+$"!), so no orientation data is really needed to integrate. This is analogous to the fact that no orientation data is needed to integrate an even 0-form over a 0-submanifold (i.e. to add up the values of a function at a set of points!).

The upshot is that a volume form or other nice measure is naturally an odd form, because the volume/measure of a submanifold certainly doesn't depend on any choice of orientation. Whereas when you try to treat volume as integration over an even form, you have to fix an orientation arbitrarily.

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  • $\begingroup$ Does this mean that if I use $\alpha = a_x dy\wedge dz + a_y dz\wedge dx + a_z dx\wedge dy$ then I am integrating an even $2$-form, but if I use $\beta = \star (a_x dx + a_y dy + a_z dz)$ then I am integrating an odd $2$-form? Because to apply the Hodge star $\star$, I need to specify whether $dx\wedge dy \wedge dz$ is positively oriented or whether $-dx\wedge dy \wedge dz$ is, from where I get the transverse orientation also. What do you mean by canonical choice for an odd $n$-form? Having to pick plus in $\star 1 = \pm dx\wedge dy \wedge dz$? Why would the $m<n$ case be non-canonical? $\endgroup$
    – lightxbulb
    Oct 28, 2023 at 5:11

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