Integration on an orientable differentiable nmanifold is defined using a partition of unity and a global nowhere vanishing nform called volume form. If the manifold is not orientable, no such form exists and the concept of a density is introduced, with which we can integrate both on orientable and nonorientable manifolds. My question is: On a nonorientable nmanifold, every nform vanishes somewhere, but shouldn't I be able to chose an nform with say a countable number of zeros, which would then constitute a set of measure zero and thus allow me to use nforms (with zeros) for global integration also on nonorientable nmanifolds?
6 Answers
The problem is that there is no way to figure out signs  It would be like trying to integrate a function from $\mathbb{R}$ to $\mathbb{R}$ without knowing whether you were moving forward or backward.
What you CAN actually integrate are pseudodifferential forms. The whole point of choosing an orientation is to turn a differential form into a psuedodifferential form. For those, I recommend the wonderful short story by John Baez found here:
https://groups.google.com/group/sci.physics.research/msg/3c6a1a7237b66c8c?dmode=source&pli=1

2$\begingroup$ Some of us find scrolling down more of a chore than (not) clicking ;) $\endgroup$ Mar 9, 2012 at 16:36

$\begingroup$ Steven, I did click, I was mildly amused though not fired with the zeal of the convert, and now I have to scroll down ever tabernac time I want to read the answer below yours :( $\endgroup$ Mar 10, 2012 at 19:45

2$\begingroup$ Since you seem to feel quite strongly about it, I guess I will roll it back out of respect (I really have enjoyed a lot of your answers on this site!). Hopefully this comment thread doesn't obstruct your scrolling pleasures when we are done with it  maybe we should delete all these comments too. $\endgroup$ Mar 12, 2012 at 6:12

2$\begingroup$ More than 10 years after, I keep using this answer as a relay station to that short story. Great answer and legendary story. $\endgroup$ Oct 7, 2023 at 21:24

2$\begingroup$ For future visitors: in case the link ever rots, note that the full text of the story is inside the edit/revision history of the answer. $\endgroup$– D.R.Mar 21 at 15:43
This is not an answer, but on reading the discussion I thought that it would be nice if someone gave the definition of a density so that no one would think that it is a complicated object. I learned the following (somewhat more general definition) from I.M. Gelfand:
Definition.
A $k$density on a manifold $M$ is a continuous realvalued function defined on the
cone of simple (a.k.a. decomposable) tangent $k$vectors on $M$ that is homogeneous of degree one. A $k$density $\varphi$ is said to be smooth if for every $k$tuple of smooth linearly independent vector fields $X_i$ $(1 \leq i \leq k)$ defined in some open set $U \subset M$, we have that the function
$$
y \mapsto \varphi(X_1(y)\wedge \cdots \wedge X_k(y))
$$
is smooth in $U$.
A densitiy is called even if $\varphi(v) = \varphi(v)$ for every simple tangent $k$vector $v$. Likewise, we have odd $k$densities that generalize differential $k$forms
Examples and context
If $\Omega$ is a volume form on a manifold of dimension $n$, then both $\Omega$ and $\Omega$ are $n$densities. The arclength element of a Riemannian or Finsler metric is a $1$density. The $k$area integrand of a Riemannian or Finsler manifold is a $k$density.
Parametric integrands (in the sense of FedererFleming) define $k$densities when restricted to the cone of simple vectors, but densities are way more general.
Varifolds of dimension $k$ are elements of the dual to the space of even $k$densities. This is basically their definition: because of their homogeneity even $k$densities can be seen as continuous functions on the bundle of tangent $k$planes.
A message from our sponsor
For more examples and for neat applications to integral geometry (if I may say so myself ...), which becomes much easier if differential forms are replaced by densities, see the paper Gelfand transforms and Crofton formulas.

1$\begingroup$ I would just like to add a word of endorsement for anything written by Juan Carlo AlvarezPaiva. $\endgroup$ Mar 9, 2012 at 16:12

4$\begingroup$ Your terminology of "kdensities" nicely clashes with 1/2densities used in geometric quantization. There are really two indices attached to densities: one is the number of vectors they eat. The other deals with how they transform, that is, a character of GL(n), which can be identified with a nonzero complex number. $\endgroup$ Feb 12, 2013 at 18:00

1$\begingroup$ The point being, that it might be good to warn people about the conflict when you use the term, that's all. $\endgroup$ Mar 1, 2013 at 21:53

2$\begingroup$ A curious fact is that Guillemin and Sternberg dedicate their book to Gelfand (from whose work they learned integral geometry, I assume) and then they define $\alpha$densities without saying their notation "nicely clashes" with Gelfand's ;) $\endgroup$ Mar 5, 2013 at 10:06

1$\begingroup$ For the benefit of people who might want to find the Gelfand and Gindikin reference, I believe it is “Nonlocal inversion formulas in real integral geometry”, Funct. Anal. Its Appl. (1977) 11, 173–179, originally in Russian in Функц. анализ и его прил. (1977) 11(3), 12–19. $\endgroup$ Jul 10, 2016 at 2:51
First of all, the things that you actually integrate are densities, which are the differential geometric counterparts of measures. No orientation is needed. On a smooth manifold $M$ with separable topology there is an intrinsic concept of negligible set. A density is then a signed measure $\mu$ such that $\mu(B)=0$ for any negligible Borel subset.
A degree $n$ form on an $n$dimensional manifold is almost a density, but not quite. We need an orientation to associate to the top degree form a density. This is what you ultimately integrate when you integrate a form. For more details see Section 3.4.1 of these notes.

$\begingroup$ Thank you for the link. I guess what I am asking is if I could forget about densities and, with my above argument just resort to nforms also in the nonorientable case. It is just that forms appear all over the place and up to now, I've seen densities used only for integration on nonorientable manifolds. So, is there no way I could get around having to use densities? $\endgroup$– ISHMar 7, 2012 at 14:25

8$\begingroup$ If integration is your goal, then you cannot avoid densities. as a matter of fact it is easier to work with densities then with forms. There is a concept of pseudoform which a bit tricky; see the book The Geometry of Physics by Theodore Frankel. $\endgroup$ Mar 7, 2012 at 14:57

7$\begingroup$ Following up on pseudoforms: A degree$n$ pseudoform on an ndimensional manifold is the same thing as a density (and hence the same thing as an absolutely continuous Radon measure) so can be integrated directly, while a degree$k$ pseudoform for $k < n$ can be integrated only with the help of a pseudoorientation (on the region of integration). Flux is a good example of the integral of a pseudoform (of degree $n  1$); the pseudoorientation specifies in which direction one is passing through. $\endgroup$ Mar 1, 2013 at 22:01
You would expect the zero set of an $n$form to have codimension 1 rather than being countable. Your suggestion of choosing some $n$form on a nonorientable manifold $M^n$ and defining integrals relative to that essentially amounts to cutting $M$ into two orientable pieces along a codimension 1 submanifold, choosing an orientation on each, and adding the integrals on the two pieces. You can certainly do that, but since the answer depends on the choice of $n$form/cutting it is not very natural or interesting (whereas the integral on an oriented manifold only depends on the orientation and not on the choice of orientation form).

$\begingroup$ « whereas the integral on an oriented manifold only depends on the orientation and not on the choice of orientation form ». Maybe you need to normalize the volume to $1$? By linearity, for $\lambda >0$, we have $\int f \cdot\lambda\cdot\mathcal{V}=\lambda\cdot\int f\cdot\mathcal{V}$, so the integral does depend on the volume form. $\endgroup$– QfwfqMar 7, 2012 at 18:40

3$\begingroup$ By integration, I mean (and interpreted the question as referring to) a linear functional on (compactly supported) $n$forms, rather than on functions. This functional depends only on the orientation. $\endgroup$ Mar 7, 2012 at 19:20

In fact, for the purpose of integrating functions on a nonorientable manifold, you don't directly need densities. Every (connected) manifold $M$ has an orientable double cover (connected if and only $M$ is not orientable), $\pi: \tilde M \to M$. Then, upon fixing an orientation and a volume form $\mathcal V$ on $\tilde M$, you can define the integral of a function $f$ on $M$ by
$$\frac{1}{2}\;\int_{\;\tilde M}\left(f\circ\pi\right) \mathcal V\,.$$ Of course, if $M$ is not orientable, the volume form $\mathcal V$ cannot be chosen so as to be invariant under the involution that swaps the two points in each fiber of $\pi$, otherwise it would be the pullback of a volume form on $M$.
This has the disadvantage of not being very canonical in general. But, if $M$ is equipped with a Riemannian metric $g$, then there is a canonical volume form on $\tilde M$, namely that induced by $\pi^* g$. So, for instance, the volume of a Riemannian Moebius strip or of a Riemannian Klein bottle is well defined.

$\begingroup$ Actually this integral comes from the theory of the Lebesgue integral which is based on the measure. $\endgroup$– C.F.GNov 13, 2020 at 22:08

$\begingroup$ 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? $\endgroup$ Oct 24, 2023 at 17:06
I think one reason that integration of forms instead of densities is prefered is that one can use Stokes theorems.


3$\begingroup$ I don't know why people voted this down? That is of course the point. However, it is not a matter of "preference" : arclength or area are not differential forms. Many things are densities in a natural way and cannot be made into differential forms. $\endgroup$ Jun 13, 2012 at 22:38

4$\begingroup$ Stokes's Theorem applies equally well to pseudoforms, which includes densities. (I mean those densities that ISH was asking about, $n$densities in the Gelfand–Gindikin sense, $1$densities in the Guillemin–Sternberg sense, with both senses also allowing more general notions of density that are not pseudoforms.) $\endgroup$ Mar 1, 2013 at 22:10

$\begingroup$ For example, if $B$ is a ball with boundary sphere $S$, this induces a pseudoorientation (notion of inside and outside) on $S$, so we can integrate pseudoforms on it. Let $\omega$ be a density (say the density of some physical substance). If we also have a vector field $v$ (say, a velocity field), then the interior product $\iota_v \omega$ is an $(n1)$pseudoform (where $n$ is the dimension of the ambient manifold). Then $\int_S \iota_v \omega = \int_B \mathrm{d}(\iota_v \omega)$, by Stokes's Theorem. … $\endgroup$ Mar 1, 2013 at 22:14

$\begingroup$ … The integral $\int_S \iota_v \omega$ is the flux of the substance through the surface $S$; if the substance is conserved, then we have $\int_S \iota_v \omega + \int_B \dot\omega = 0$, where the dot indicates differentiation with respect to time. (That is, imagine spacetime as the cartesian product of a space manifold with a time line; our fields are defined on all of spacetime but are thought of as forms only on space.) Since $B$ could be any ball, conclude that $\mathrm{d}(\iota_v \omega) + \dot\omega = 0$, the differential equation of continuity. $\endgroup$ Mar 2, 2013 at 20:33