Integration on manifolds

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1. Mar 15, 2015

"Don't panic!"

In all the notes that I've found on differential geometry, when they introduce integration on manifolds it is always done with top forms with little or no explanation as to why (or any intuition). From what I've manage to gleam from it, one has to use top forms to unambiguously define integration on a manifold (although I'm not quite sure why this is the case?!) and one can integrate lower dimensional forms via integration on a chain (through defining pullbacks). I'm really struggling to understand these notions, please could someone enlighten me on the subject?

2. Mar 16, 2015

Bacle2

Well, one issue is the independence of integration of forms on the choice of coordinates, which is not the case for functions.

3. Mar 16, 2015

lavinia

What is your reference? Can you illustrate what you are talking about?

k-forms can be integrated over smooth k-chains. These do not need to be top dimensional forms.
In fact, the entire cohomology with real coefficients can in principal be computed from integrals of k-forms over smooth k-chains..

4. Mar 16, 2015

"Don't panic!"

The notes that I've managed to find all introduce integration on manifolds with a phrase like "consider an n-form defined on an n-dimensional manifold. The integral of such an n-form is...". They also state things such as "n-forms are natural objects to integrate on a manifold as they do not require a metric". I was trying to find some motivation for these things and thought that it might have something to do with orientability issues, especially after reading this sentence on the Wiki page: "There is in general no meaningful way to integrate k-forms over subsets for because there is no consistent way to orient k-dimensional subsets".

The issue arose as myself and a colleague were trying to figure out the following expression: $$V=\int_{0}^{R}dr\;4\pi r^{2}\sqrt{1-\left(\frac{dt}{dr}\right)^{2}}$$ where $t=t(r)$.

Which is apparently the volume enclosed by a 3-dimensional sphere (of radius $R$) in Minkowski spacetime?! We wondered why one couldn't just integrate over 3-dimensional space using a 3-D volume element and assumed that one had to integrate the 4-volume form for Minkowski space, i.e. $$\int dV= \int\sqrt{-g}dt\wedge dx\wedge dy \wedge dz$$ with the standard orientation $+(t,x,y,z)$, and introduce a pull-back map to constrain one of the degrees of freedom to obtain a 3-dimensional volume integral, but weren't sure as to why (or if our intuition was correct)?!

Last edited: Mar 16, 2015
5. Mar 16, 2015

lavinia

The reference starts with the assumption that you want to integrate something over the entire manifold and states that n-forms are the natural candidates.
But to do this one needs to in principal express the manifold as a n-chain and then piece a global form together using partitions of unity subordinate to the smooth simplices in the n chain. The fundamental object is the n-simplex. But one could have a k-simplex k<n and integrate a k form over it. The idea is exactly the same.

I do not know relativity theory but the formula seems to be a simplification of the general volume integral for the special case of a sphere. What is meant by a sphere in Space-Time?

Last edited: Mar 16, 2015
6. Mar 16, 2015

Ben Niehoff

You can integrate k-forms over k-chains by pulling them back, at which point they become top forms that live on the k-chain itself. So you really only need to define integration of top forms, and pullbacks. My understanding of chains is that a chain is not just a set, but also has an orientation, so the integration of forms is well-defined.

Calculating volumes and areas is trickier, though. Consider the simpler case of calculating arc lengths. You know what the arc length functional is:

$$\int_\gamma d\lambda \, \sqrt{ \pm g_{\mu\nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda} }$$
You can think of this as integrating the pullback along $\gamma$ of the form

$$\sqrt{ \pm g_{\mu\nu} \, dx^\mu \, dx^\nu }$$
But this object is not a 1-form, because it is not a linear map from $T_xM \to \mathbb{R}$. In general, there is no linear form whose pullback along $\gamma$ gives the arc length functional for all possible $\gamma$.

A similar situation applies to area functionals of any k-area for $k < n$. These area functionals cannot be linear forms, but are more complicated objects (essentially, square roots of various determinants). The exception is the top form on $M$, which is linear only because $\Lambda^n T_xM$ is 1-dimensional.

Last edited by a moderator: Apr 19, 2017
7. Mar 16, 2015

"Don't panic!"

So is it possible to directly integrate a k-form (with k<n) on an n-dimensional manifold?

The Minkowski space-time is a 4-dimensional manifold with zero curvature (by space-time in general it is meant that one is considering a 4-dimensional Pseudo-Riemannian manifold, with 3 spatial coordinates and 1 temporal coordinate). Basically, I think they are integrating a 3-d sphere on such a manifold, but I don't see where the expression I gave above comes from?!

8. Mar 16, 2015

"Don't panic!"

Ah, so is the reason that integration on manifolds seems to be introduced in terms of top forms because these are the natural objects to integrate on an n-dimensional manifold, and then by introducing the notion of a pull-back map one can always map a top-form to a lower dimensional form [k(<n) form] and integrate this on a submanifold?

9. Mar 16, 2015

lavinia

Sure. What about a line integral around a closed curve in the plane.? This is a 1-form integrated over a 1 chain in a two dimensional manifold.

10. Mar 16, 2015

lavinia

No. What Ben was saying was the a k- form is integrated over a smooth k-simplex so with respect to the k-simplex it is automatically top dimensional. So you only need to define the integrals of forms in the top dimension.

When one pulls a k-form back along a k-simplex one ends up integrating over the standard k-simplex in Euclidean space. This is a k-manifold with boundary. For a k-manifold without boundary, one must in principal express the manifold as a linear combination of k-simplces with boundary components cancelling.

11. Mar 16, 2015

"Don't panic!"

So is the point that one defines integration for top forms on manifolds and then uses pullback maps to create chains to reduce it to a lower dimensional integral, e.g. for the example you gave would one pullback the 1-form defined on the two dimensional manifold to a 1-form defined on the one-dimensional sub-manifold? (for example, if $\omega\in\Omega^{1}(\mathbb{R}^{2})$ and $\phi :\mathbb{R}\rightarrow\mathbb{R}^{2}$, then $\phi^{\ast}\omega\in\Omega^{1}(\mathbb{R})$ such that $$\int_{\phi}\omega =\int_{\mathbb{R}}\phi^{\ast}\omega\;\;)$$

12. Mar 16, 2015

lavinia

No. ω is defined on the plane but its integrals are defined on 1 chains. One can not integrate it over $R^2$

13. Mar 16, 2015

"Don't panic!"

Sorry, I noticed my original error, but wasn't able to correct it quickly enough, is the subsequent correction I made correct? Is it correct to say that the chain is given via the map $\phi :\mathbb{R}\rightarrow\mathbb{R}^{2}$ such that $\omega$ is integrated over the chain $\phi$,i.e. $$\int_{\phi}\omega =\int_{\mathbb{R}}\phi^{\ast}\omega$$

14. Mar 16, 2015

lavinia

Yes except that a 1 simplex is an oriented line segment with end points included. It is not all of R.

15. Mar 16, 2015

"Don't panic!"

Can one extend this to arbitrary (but finite) dimension, i.e. given an n-form $\omega\in\Omega^{n}(M)$ defined on some n-dimensional manifold $M$, then one can define a set of $k$-chains such that one can pull-back the n-form $\omega$ to a k-form defined on some k-dimensional sub-manifold $N$ of $M$ and then integrate this over a subset of $N$. That is (something of the form) $$\int_{\phi (D)}\omega =\int_{D}\phi^{\ast}\omega$$ where $D\subset N$?!

16. Mar 16, 2015

lavinia

The pull back of a n-form will be an n-form not a k form. The process of pulling back does not change the dimension.

Last edited: Mar 16, 2015
17. Mar 16, 2015

"Don't panic!"

yes, sorry, you're right. It's the manifold that we are considering that changes dimension, right?

18. Mar 16, 2015

lavinia

Not really. It is true that one starts with a n-manifold and ends up integrating on a k manifold. But one is not integrating over the n-manifold itself so you are not changing dimension. One is integrating over a k-dimensional simplex inside the manifold.

19. Mar 16, 2015

"Don't panic!"

Excuse my ignorance, but what is a "simplex". Are we integrating over a k-dimensional submanifold of the original manifold?

20. Mar 16, 2015

lavinia

A simplex is a smooth mapping of the standard simplex in Euclidean space into the manifold. One integrates it by pulling it back to the standard simplex in Euclidean space. A simplex is not a submanifold. It is a mapping. I apologize because I feel that I have confused things by bringing in the idea of simplexes.

Let's go back the case of a line integral. Here one integrates along a parameterized curve. The parameterization is a mapping of a closed interval into the plane.

If there is a 1-form defined on the plane one can pull it back and then integrate it over the interval. This is the same as integrating the function you get by evaluating the 1-form on the velocity vectors of the curve.

In higher dimensions you need to generalize the idea of an oriented interval and usually this is done formally by taking a standard set of oriented triangles, tertrahedra and their higher dimensional analogues. These are called standard simplexes. If you like you could instead use oriented squares and cubes and hypercubes or something similar. In Mathematics these these are all used but usually the formal definition is in terms of standard simplexes. Though I have seen papers where cubes are more convenient.

The terminology is a little confusing because a standard simplex is an oriented geometric object in Euclidean space while a smooth simplex is a smooth mapping of the standard simplex into a manifold.

Last edited: Mar 16, 2015