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I am trying to understand, in as simple terms as possible:

  1. How to define integration for non-orientable manifolds, and
  2. why it is impossible to do so using only differential forms.

In particular, I've seen some discussion of using "densities" instead of $n$-forms for integration, but am not really clear on why densities are required. In other words, is it really impossible to define integration on nonorientable manifolds using forms alone?

I am of course aware that any $n$-form must vanish somewhere on a nonorientable manifold, so we cannot find a volume form, hence cannot use the standard definition of integration. I think the reason I'm not finding this answer satisfying is that it is a bit tautological: we can't define integration with respect to volume forms because there are no volume forms. But why must we define integration with respect to a (global) volume form in the first place? Is there really no other way to do it using locally-defined forms? Thinking of a manifold as a collection of local charts is common in geometry, and I'm having trouble understanding why this approach doesn't work in the case of integration.

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If you are defining an orientation as a class of $n$-forms for the equivalence relation given by multiplication by positive functions, then there are waaaay too many orientations. –  Mariano Suárez-Alvarez Oct 5 '10 at 18:19
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Why do regard this positive number you get as a "volume"? –  Robin Chapman Oct 5 '10 at 18:39
    
I didn't understand how you're going to do this procedure: "We can then use our partition of unity to "sum up" these positive values over the manifold to get a positive total volume." –  Ronaldo Oct 6 '10 at 2:18
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The reason why people define integration with respect to forms is because they want a situation where you can generalize the fundamental theorem of calculus -- which is implicitly an oriented concept, as the interval has an initial point and a terminal point. That generalization is Stokes' theorem. There are of course all kinds of other notions of integration and they're all perfectly fine. But you use forms when you want to integrate with respect to oriented volumes, not just plain old measures. –  Ryan Budney Oct 6 '10 at 17:33
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Reading your question again, I notice you don't specify what you want to integrate. If you're interested in integrating real-valued functions then densities are precisely what you need. But if you're happy integrating other things (like differential forms) then differential forms are all you need. –  Ryan Budney Apr 21 '11 at 23:30

1 Answer 1

On an orientable manifold, we define integration of functions with respect to a volume form. On a non orientable manifold, there are no volume forms, so we have to do something else!

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Thanks for your answer. I think the reason I'm not finding it satisfying is that it's a bit tautological: we can't define integration with respect to volume forms because there are no volume forms. But why must we define integration with respect to (global) volume forms? Is there really no other way to do it using local forms? Thinking of a manifold as a collection of local charts is common in geometry, and I'm having trouble understanding why it's impossible in the case of integration. –  funarharpsichord Oct 6 '10 at 15:46
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@funarharpsichord: well, what do you want to do with the integral on a non-orientable manifold? The reason we define integration on manifolds as we do is because it is the only way to do the things we want to do. Maybe you could state a concrete problem that you want to solve, and then we may be more concrete. –  Mariano Suárez-Alvarez Oct 6 '10 at 17:09
    
Forms are needed because integrals have a sign. For instance the sign of a line integral depends on the direction that the curve is parameterized. In general this idea leads to the idea of an orientation of the manifold. I you just want to compute unoriented areas, you do not strictly speaking need differential forms. I wonder if you could generaze integration of functions to unorientable manifolds by fulling the function back to the 2 fold cover, integrate there, then divide by 2. –  lavinia Dec 13 '13 at 17:27

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