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Integration on an orientable differentiable n-manifold is defined using a partition of unity and a global nowhere vanishing n-form 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 non-orientable manifolds. My question is: On a non-orientable n-manifold, every n-form vanishes somewhere, but shouldn't I be able to chose an n-form with say a countable number of zeros, which would then constitute a set of measure zero and thus allow me to use n-forms (with zeros) for global integration also on non-orientable n-manifolds?

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5 Answers 5

up vote 17 down vote accepted

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 pseudo-differential forms. The whole point of choosing an orientation is to turn a differential form into a psuedo-differential form. For those, I recommend the wonderful short story by John Baez found here:


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Some of us find scrolling down more of a chore than (not) clicking ;) –  Yemon Choi Mar 9 '12 at 16:36
But only those who would not have clicked - exactly the audience I was trying to reach! –  Steven Gubkin Mar 10 '12 at 1:19
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 :( –  Yemon Choi Mar 10 '12 at 19:45
Is it really too much to ask that you could editorialize? –  Yemon Choi Mar 10 '12 at 19:47
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. –  Steven Gubkin Mar 12 '12 at 6:12

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 real-valued 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 arc-length 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 Federer-Fleming) 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.

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I would just like to add a word of endorsement for anything written by Juan Carlo Alvarez-Paiva. –  Deane Yang Mar 9 '12 at 16:12
Your terminology of "k-densities" nicely clashes with 1/2-densities 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. –  Eugene Lerman Feb 12 '13 at 18:00
It's not my terminology. It is Gelfand and Gindikin's, I believe, and I think they introduced it before Guillemin and Sternberg used 1/2 densities. –  alvarezpaiva Feb 14 '13 at 20:57
When I first read your answer at mathoverflow.net/questions/99488/… claiming that the arclength element in a Riemannian manifold is a 1-density, I thought that this was wrong, since I knew the notation mentioned by Eugene. (In that notation, a 1-density is the same thing a top-rank pseudoform, so I took your answer at first to mean that ds is a pseudoform on the curve seen as a 1-dimensional manifold in its own right, which is not exactly wrong but also not good enough.) –  Toby Bartels Mar 1 '13 at 21:52
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 ;-) –  alvarezpaiva Mar 5 '13 at 10:06

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 non-orientable 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).

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« 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. –  Qfwfq Mar 7 '12 at 18:40
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. –  Johannes Nordström Mar 7 '12 at 19:20
Ok . –  Qfwfq Mar 7 '12 at 20:19

First of all, the things that you actually integrate are densities, which are the differential geometric counterparts of measures. No orientation is needed.

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 page 105 of these notes.

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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 n-forms also in the non-orientable case. It is just that forms appear all over the place and up to now, I've seen densities used only for integration on non-orientable manifolds. So, is there no way I could get around having to use densities? –  ISH Mar 7 '12 at 14:25
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 pseudo-form which a bit tricky; see the book The Geometry of Physics by Theodore Frankel. –  Liviu Nicolaescu Mar 7 '12 at 14:57
Following up on pseudo-forms: A degree-$n$ pseudo-form on an n-dimensional 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-$k$ pseudo-form for $k < n$ can be integrated only with the help of a pseudo-orientation (on the region of integration). Flux is a good example of the integral of a pseudo-form (of degree $n - 1$); the pseudo-orientation specifies in which direction one is passing through. –  Toby Bartels Mar 1 '13 at 22:01

I think one reason that integration of forms instead of densities is prefered is that one can use Stokes theorems.

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Are you sure about that? –  Pait Mar 15 '12 at 12:19
I don't know why people voted this down? That is of course the point. However, it is not a matter of "preference" : arc-length or area are not differential forms. Many things are densities in a natural way and cannot be made into differential forms. –  alvarezpaiva Jun 13 '12 at 22:38
Stokes's Theorem applies equally well to pseudo-forms, 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 pseudo-forms.) –  Toby Bartels Mar 1 '13 at 22:10
For example, if $B$ is a ball with boundary sphere $S$, this induces a pseudo-orientation (notion of inside and outside) on $S$, so we can integrate pseudo-forms 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 $(n-1)$-pseudo-form (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. … –  Toby Bartels Mar 1 '13 at 22:14
… 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. –  Toby Bartels Mar 2 '13 at 20:33

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