Foliation of 4-dimensional connected Hausdorff orientable paracompact manifold

1. Aug 3, 2011

ivl

Hi all!
I am reading a book on Classical Electrodynamics (Hehl and Obukhov, Foundations of Classical Electrodynamics, Birkhauser, 2003). In this manual I found the following statement:

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Consider a 4-dimensional differentiable manifold which is:
-connected (every 2 points are connected by continuous curve)
-Hausdorff (http://en.wikipedia.org/wiki/Hausdorff_space#Definitions)
-orientable (http://en.wikipedia.org/wiki/Orientability#Orientation_of_differential_manifolds)
-paracompact (manifold covered by finite number of coordinate charts)
This manifold always has a foliation by 3-dimensional hypersurfaces (each hypersurface is a hypersurface of constant "time").
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Does anyone know another reference which confirms this statement? In other words, do you know why a connected Hausdorff orientable paracompact manifold always has such foliation?

Any help is massively appreciated!
Cheers!

2. Aug 3, 2011

holy_toaster

Hi there!

I have read that book some time ago and remember to have struggeled with this claim, too. I have come to the conclusion that this assertion is not true in this generality. At least not globally. But I think the authors do not even want to have such a strong global statement, because for developing electrodynamics essentially all you need is that the manifold (i.e. the spacetime in this case) splits locally (in some open set) as a product of the reals and a 3-dim hypersurface.

Furthermore the problem is not really well posed in the global case. As you have not only any 4-dimensional manifold but a 4-dim spacetime with a Lorentzian metric the timelike character of one dimension adds further problems, that hinders one to use usual splitting theorems from Riemannian geometry. Because of the additional assumption that you want to have the 3-dimensional hypersurfaces as slices of constant time, you essentially demand that the 4-dim spacetime (say $M$) has a product structure $M=\mathbb{R}\times S$ with $S$ being some Riemannian metric.

For a spacetime you can only really assure this global splitting if it is http://en.wikipedia.org/wiki/Globally_hyperbolic" [Broken]

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3. Aug 3, 2011

zhentil

If one also demands that the hypersurfaces be orientable, it would imply that the manifold has euler characteristic zero (which does not strike one as the "general" case).

4. Aug 3, 2011

zhentil

The existence of such a foliation does not imply that the manifold is a trivial product. Any three manifold crossed with S^1 has such a structure.

5. Aug 3, 2011

zhentil

Other obstructions: if M is simply connected and closed, no such structure can exist.

6. Aug 4, 2011

holy_toaster

As the manifold in question should be a Lorentzian manifold, especially a spacetime, things are a bit more complicated than just asking if there exists any 3-dim foliation. This foliation should also be transverse to the time direction. Meaning the leaves should be spacelike slices.

So if the manifold is compact (although compactness is not a very natural condition for spacetimes because any compact spacetime has closed timelike curves) it must have Euler characteristic zero to allow for the global existence of a Lorentzian metric in the first place.

If the manifold is not compact and you have not much other choice than demanding global hyperbolicity. You then get a foliation by 3-dim Cauchy hypersurfaces and a trivial product structure.

7. Aug 4, 2011

ivl

Hi all,
thanks for the useful replies
===========
@holy_toaster:
I checked the paper by Mielke and Wallner, from which Hehl and Obukhov take their foliation (see: Mielke and Wallner, Il Nuovo Cimento B, 101B, 1988). Mielke and Wallner take the manifold M to have structure:
M=RxN
where the N is a 3-dimensional manifold. I suspect, then, that the book by Hehl and Obukhov also meant M=RxN (do you agree?)

The choice of M=RxN in Hehl and Obukhov was then criticised by Delphenich (Ann. Phys. (Leipzig) 14, No. 6, 347 – 377 (2005)). Probably in the light of this criticism, the 2006 article by Hehl and Obukhov (Lecture Notes in Physics, 2006, Volume 702/2006, 163-187) says that the foliation is applied locally.

The questions now are:
1. A recent paper by Dahl (http://arxiv.org/abs/1103.3118) also takes the manifold to be M=RxN. Maybe M=RxN is a restrictive assumption, but an assumption that works well nonetheless? Can we do pre-metric electrodynamics starting with M=RxN?

2. The idea of a "local" foliation implies no restrictions. Indeed, any manifold has local foliations by means of the local coordinates. The idea of a "local"foliation is perhaps not restrictive enough... what do you reckon?

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@zhentil:
Do the assumptions:
-connected (every 2 points are connected by continuous curve)
-Hausdorff (http://en.wikipedia.org/wiki/Hausdor...ce#Definitions [Broken])
-orientable (http://en.wikipedia.org/wiki/Orienta...tial_manifolds [Broken])
-paracompact (manifold covered by finite number of coordinate charts)
not imply that the Euler characteristic is zero? Do they not imply that all closed p-forms are also exact?

============

Thanks a lot!

Last edited by a moderator: May 5, 2017
8. Aug 4, 2011

holy_toaster

Yes, I think Hehl and Obukhov meant a global product structure $M=\mathbb{R}\times N$ in the book, such that the 3-dim hypersurfaces of constant $t\in\mathbb{R}$ are spacelike everywhere. Maybe they believed that this was always possible. I wasn't aware of the articles that followed and questioned this assumption.

Let me go more into detail about the splitting problem: if you have a global timelike vector field $V$ on your Lorentzian manifold, then under quite mild assumptions on $V$ this leads to a product structure $M=\mathbb{R}\times N$, where $V$ points in the direction of growing $t\in\mathbb{R}$. This means you have 3-dim hypersurfaces transversal (not necessarily perpendicular!) to $V$. (This problem is part of my PhD thesis, so let me know if you want to have more details about it.)

But the problem is that for doing a field theory like electrodynamics, this may be not enough indeed, because the 3-dim leaves of the foliation are not necessarily everywhere spacelike. This especially happens if there are causality violations. For example take the http://en.wikipedia.org/wiki/G%C3%B6del_metric" [Broken]. Its manifold is just $\mathbb{R}^4$ but no matter what foliation by 3-dim hypersurfaces transverse to a timelike vector field you choose, the leaves are always non-spacelike somewhere. This has to do with the fact that there is a CTC through every point of Goedel spacetime.

Thus for having globally spacelike leaves of the foliation there must be involved some causality condition on the Lorentzian metric. I think in general stable causality suffices to ensure this, but one often considers global hyperbolicity because it has some additional good properties ("[URL [Broken]

1. I do not think that the product structure $M=\mathbb{R}\times N$ is too restrictive. I think it is a quite natural assumption for a spacetime on which you want to do some physical field theory, because it distinguishes some global time coordinate $t\in\mathbb{R}$. I do see no problem to do the Hehl/Obukhov premetric electrodynamics with the assumption of this structure of the manifold.

2. Now when I think about it again I am not completely sure any more that the local foliation by coordinates is really enough for this premetric electrodynamics. It was some time ago when I was doing this stuff (so correct me if i am mistaken here) but I think I recall that there are not only tensors involved, but also some intergals over compact regions from the very beginning to define things like the electric charge. So maybe what one should have at least is a foliation by spacelike hypersurfaces in any compact subset. This is a weaker causality condition than stable causality and was first investigated by Beem I think in: Beem, Conformal Changes and Metric Completeness, Commun. Math. Phys., Vol. 49, p.179, (1976).

Now that I am writing this the thought crosses my mind that all this is maybe not what you are looking for in relation to premetric electrodynamics...
Because (correct me if I am mistaken) the whole idea of this theory is to start out with some manifold without a metric, construct on this manifold an electrodynamic theory by some reasonable axioms and then reconstruct the metric or at least its conformal class from this electrodynamic theory...
But nevertheless the product structure $M=\mathbb{R}\times N$ seems to me a natural assumption for this because you need a priori some distinguished time coordinate which is provided by $t\in\mathbb{R}$ in this case...

So maybe the question should be asked in the following way: Given a 4-dim manifold $M$ (connected, oriented, and so on) which allows for the existence of Lorentzian metrics at all, does it then allow also for a codimension 1 foliation so that there is some Lorentzian metric $g$ on $M$ such that all the leaves are spacelike with repsect to $g$? The answer to this question is (maybe surprisingly) positive: See chapter 3 in http://epub.uni-regensburg.de/20482/" [Broken] paper.

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9. Aug 4, 2011

ivl

Hi there! Thanks for the very good reply!

>Yes, I think Hehl and Obukhov meant a global product structure M=R×N in the book, such >that the 3-dim hypersurfaces of constant t∈R are spacelike everywhere. Maybe they >believed that this was always possible. I wasn't aware of the articles that followed and >questioned this assumption.

Good, we agree fully on this.

>Let me go more into detail about the splitting problem: if you have a global timelike >vector field V on your Lorentzian manifold, then under quite mild assumptions on V this >leads to a product structure M=R×N, where V points in the direction of growing t∈R. This >means you have 3-dim hypersurfaces transversal (not necessarily perpendicular!) to V. >(This problem is part of my PhD thesis, so let me know if you want to have more details >about it.)

I agree, I have read the same thing in a paper by Delphenich (same author of a paper I suggested before). I am not an expert, but I think what you say has to do with a famous paper by Geroch, concerning the Cauchy problem. Right?

>I do not think that the product structure M=R×N is too restrictive. I think it is a quite >natural assumption for a spacetime on which you want to do some physical field theory, >because it distinguishes some global time coordinate t∈R. I do see no problem to do the >Hehl/Obukhov premetric electrodynamics with the assumption of this structure of the >manifold.

Again, I agree. Even dimensionally it makes sense. In particular:
-Observe that you need to define a dimension of time (a "unit" of time in loose sense).
-However, you have no metric to help you define the dimension of time throughout the manifold (you cannot say this line is timelike, thus its dimension is time).
-Hence, the dimension of time must be defined globally by introducing a foliation M=RxN.
-By the same take, assuming a "local" foliation is no good. In fact, if you consider different foliations in different open sets, you then need to patch the foliations to get an agreeing dimension of time.

>Now when I think about it again I am not completely sure any more that the local >foliation by coordinates is really enough for this premetric electrodynamics. It was some >time ago when I was doing this stuff (so correct me if i am mistaken here) but I think I >recall that there are not only tensors involved, but also some intergals over compact >regions from the very beginning to define things like the electric charge. So maybe what >one should have at least is a foliation by spacelike hypersurfaces in any compact subset. >This is a weaker causality condition than stable causality and was first investigated by >Beem I think in: Beem, Conformal Changes and Metric Completeness, Commun. Math. >Phys., Vol. 49, p.179, (1976).

Hmm... interesting... On a slightly different topic, it turns out that:
-Delphenich criticises Hehl+Obukhov for using M=RxN (again: Annalen der Physik, 14: 347–377, 2005)
-Yet, in the same paper, Delphenich himself says "There are reasons to suspect that this situation is based in deeper considerations about the nature of wave motion. As has been discussed elsewhere [17], an essential construction that is associated with wave motion in a four-dimensional manifold is that of a pair of transverse foliations of M of codimension one. One of them defines the proper time simultaneity leaves relative to a choice of rest frame. The other defines the isophase hypersurfaces for the wave motion. Their intersections are two-dimensional instantaneous isophase surfaces."
-So, in the end, Delphenich needs the foliation of constant time anyway. Notice that the second foliation (constant electromagnetic phase) is also treated correctly in Hehl+Obukhov.

>Now that I am writing this the thought crosses my mind that all this is maybe not what >you are looking for in relation to premetric electrodynamics...
>Because (correct me if I am mistaken) the whole idea of this theory is to start out with >some manifold without a metric, construct on this manifold an electrodynamic theory by >some reasonable axioms and then reconstruct the metric or at least its conformal class >from this electrodynamic theory...
>But nevertheless the product structure M=R×N seems to me a natural assumption for this >because you need a priori some distinguished time coordinate which is provided by t∈R in >this case...

Your last comment is illuminating... see my previous comment about the need to define a dimension of time.
Thanks for the nice discussion!

10. Aug 4, 2011

lavinia

Because the orientations of the hypersurfaces can be alligned with each other in a coordinate chart in which the hypersurfaces are coordinatized as hyperplanes?

11. Aug 4, 2011

lavinia

How do you foliate the 4 sphere?

12. Aug 4, 2011

zhentil

the hyperplane distribution is complemented by a real line bundle. It is orientable if and only if the distribution is. If it is orientable it's trivial, yielding a nonvanishing vector field.

13. Aug 4, 2011

zhentil

To the original poster, why would that imply zero Euler characteristic? It certainly doesn't, but why should it?

14. Aug 17, 2011

ivl

You are right, no reason why it should.... !