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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.

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I suspect you are looking for a definition of 1-densities. –  Eugene Lerman Feb 7 '13 at 15:13
    
They're not densities. They're forms with coefficients in a (flat) real line bundle. –  Liviu Nicolaescu Feb 7 '13 at 15:15
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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 $\sqrt{dx^2 + dy^2}$ –  alvarezpaiva Feb 7 '13 at 15:33
    
By the way, here is another MO question on this topic mathoverflow.net/questions/99488. –  alvarezpaiva Feb 7 '13 at 15:35
    
Is your definition of a density functorial? –  Nevermind Feb 7 '13 at 16:01
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2 Answers 2

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.

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thanks, I will look on these references tomorrow, since I don't have access right now. –  Nevermind Feb 7 '13 at 16:22
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up vote 2 down vote accepted

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.

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Just using the absolute value here sounds a bit arbitrary to me. What's the reason for that and how to generalize? –  Nevermind Feb 12 '13 at 0:02
    
Just use the bundle of $k$-frames ($k$ linearly independent tangent vectors), which is a principal $GL(k)$ bundle. The rest is the same. –  Deane Yang Feb 12 '13 at 0:03
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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. –  Deane Yang Feb 12 '13 at 0:05
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@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. –  Michael Murray Feb 12 '13 at 11:09
    
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. –  Tim Campion Sep 22 '13 at 22:01
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