5
$\begingroup$

I'm looking for a definition of pseudo differential forms in ordinary differential geometry. However searching the web gave me nothing. There are definitions in supergeometry but that is not what I'm after.

Recently I read, that pseudo-differentialforms are the natural structure to integrate, since integration works on any kind of submanifold (orientation not required) for them, but those texts don't gave a 'clean' definition of these kind of forms.

What are pseudo-differentialforms?

Can pseudo differentialforms be defined as sections of some kind of fiber bundle? If yes that's a definition I would prefer.

$\endgroup$
11
  • $\begingroup$ I suspect you are looking for a definition of 1-densities. $\endgroup$ Feb 7, 2013 at 15:13
  • $\begingroup$ They're not densities. They're forms with coefficients in a (flat) real line bundle. $\endgroup$ Feb 7, 2013 at 15:15
  • 2
    $\begingroup$ I'm guessing you're looking for the notion of k-densities as explanined in my answer to this MO question: mathoverflow.net/questions/90455 If you insist on a complicated definition they are sections of a determinant line bundle over the grassmannian bundle on manifold, but they're simple objects that we use every day like $\sqrt{dx^2 + dy^2}$ $\endgroup$ Feb 7, 2013 at 15:33
  • $\begingroup$ By the way, here is another MO question on this topic mathoverflow.net/questions/99488. $\endgroup$ Feb 7, 2013 at 15:35
  • $\begingroup$ Is your definition of a density functorial? $\endgroup$
    – Nevermind
    Feb 7, 2013 at 16:01

2 Answers 2

3
$\begingroup$

There at least two sources I am aware of.

  1. Theodore Frankel, The Geometry of Physics, Section 2.8 and 3.4.

  2. Georges De Rham, Varietes Differentiables. Formes, courants, formes harmoniques, Chap. II.

$\endgroup$
1
  • $\begingroup$ thanks, I will look on these references tomorrow, since I don't have access right now. $\endgroup$
    – Nevermind
    Feb 7, 2013 at 16:22
3
$\begingroup$

Pseudo-Forms:

Let $M$ be a topological manifold and $PM$ the frame bundle of $M$. If $dim(M)=n$ then $PM$ is a $Gl(n)$-principal bundle.

Let $\tau: Gl(n) \to \mathbb{R} \; ; \; A \mapsto abs(det(A))$ the map, that maps any linear isomorphism $f \in Gl(n)$ to the absolute value of its determinant. This defines a left action of $Gl(n)$ on $\mathbb{R}$ by

$$\cdot: Gl(n) \times \mathbb{R} \to \mathbb{R} \; ; \; (A,x) \mapsto \tau(A)x$$

The bundle of pseudo-forms then is the associated (line) bundle

$$PM[\mathbb{R},\cdot]$$

of this action and pseudo-forms are sections of this bundle. If $M$ is smooth, this is a smooth bundle,since the action is smooth. ($\tau$ is smooth since $det(A)\neq0$ for $A\in Gl(n)$)

But this gives only pseudo-forms that behaves right in respect to integration on $dim(M)$-dimensional submanifolds. Remains the question, ow to generalize this to submanifolds of arbitrary dimension.

$\endgroup$
5
  • $\begingroup$ Just using the absolute value here sounds a bit arbitrary to me. What's the reason for that and how to generalize? $\endgroup$
    – Nevermind
    Feb 12, 2013 at 0:02
  • $\begingroup$ Just use the bundle of $k$-frames ($k$ linearly independent tangent vectors), which is a principal $GL(k)$ bundle. The rest is the same. $\endgroup$
    – Deane Yang
    Feb 12, 2013 at 0:03
  • 2
    $\begingroup$ You want to use the absolute value, so the integral does not depend on orientation and the thing you're integrating looks more like a measure on the manifold or submanifold, rather than a differential form. in particular, it allows you to define integration on a non-orientable manifold. If you use forms without the absolute value, there are no global sections to integrate. $\endgroup$
    – Deane Yang
    Feb 12, 2013 at 0:05
  • 1
    $\begingroup$ @Nevermind. Remember the co-ordinate change formula for the integral ? What comes out is the absolute value of the determinant of the jacobian of the co-ordinate change function. If you don't have an orientation this is how things have to transform to get a sensible definition of integral. $\endgroup$ Feb 12, 2013 at 11:09
  • $\begingroup$ From a top-down perspective, there are only so many linear actions of [the multiplicative group of R] on a 1-dimensional vector space, so that maybe it's not surprising that this one crops up here. $\endgroup$
    – Tim Campion
    Sep 22, 2013 at 22:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.