Why do I need densities in order to integrate on a non-orientable manifold?

Question

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?

Answer 1

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:

https://groups.google.com/group/sci.physics.research/msg/3c6a1a7237b66c8c?dmode=source&pli=1

Answer 2

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.

Answer 3

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.

Answer 4

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).

Answer 5

In fact, for the purpose of integrating functions on a non-orientable 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.

Answer 6

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