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

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

2012-03-07T13:52:32Z

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

2012-03-07T14:19:19Z

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.

Answer by Steven Gubkin for Why do I need densities in order to integrate on a non-orientable manifold?

2012-03-07T16:23:38Z

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

2012-03-07T16:38:44Z

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

2012-03-09T15:46:26Z

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

2012-03-11T01:52:33Z

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