I Integral subbundle of 6 KVFs gives a spacetime foliation by 3d hypersurfaces

[cianfa72]

TL;DR Summary

Spatial homogeneity and isotropy provide a spacetime foliation by spacelike 3d hypersurfaces

From this lecture at minute 15:00 onwards, the conditions for spacetime spatially homogenous and isotropic imply the existence of 6 $\mathbb R$-linear independent spacelike Killing Vector Fields (KVFs) w.r.t. the metric tensor $g$.

The lecturer (Dr. Schuller) claims that such 6 independent KVFs define spacelike hypersurfaces foliating the spacetime. Those KVFs close w.r.t. Lie bracket commutator, therefore by Frobenius's theorem they are integrable.

My question is: how one can conclude that the integral of such (integrable) subbundle actually provides a foliation by hypersurfaces of dimension 3 ?

 

[martinbn]

At each point the KVFs span a three dimensional space.

 

[cianfa72]

At each point the KVFs span a three dimensional space.

Ok surely, but the point is: why such 6 spacelike KVFs (with their commutation relations) span a subbundle of dimension 3 ?

For instance on 3d Euclidean space, the 3 $\mathbb R$-linear independent rotating KVFs span a two dimensional subbundle (the integral submanifolds of are the 2-spheres).

 

[PeterDonis]

For instance on 3d Euclidean space, the 3 $\mathbb R$-linear independent rotating KVFs span a two dimensional subbundle (the integral submanifolds of are the 2-spheres).

This is not correct: "the integral submanifolds are 2-spheres" does not mean the rotating KVFs span a 2-d subbundle. What the rotating KVFs do is define a foliation of the 3-dimensional space by 2-spheres. (Strictly speaking, you have to first pick a "center point", which picks out one particular subset of the rotating KVFs.)

 

[cianfa72]

This is not correct: "the integral submanifolds are 2-spheres" does not mean the rotating KVFs span a 2-d subbundle. What the rotating KVFs do is define a foliation of the 3-dimensional space by 2-spheres. (Strictly speaking, you have to first pick a "center point", which picks out one particular subset of the rotating KVFs.)

Why not ? The 3 independent rotating KVFs about the same "center point", when evaluated at any point/event in Euclidean space, define/span a 2-dimensional linear subspace of the tangent space at that point. Their disjoint union is a subbundle (i.e. a distribution of dimension 2). The integral submanifolds (2-spheres) of such 3 rotating KVFs actually foliate the entire 3d Euclidean space.

 

[PeterDonis]

when evaluated at any point/event in Euclidean space

Which is irrelevant in this discussion because we are talking about the entire Killing vector fields, not just the vectors they evaluate to at a particular point and the integral curves those vectors give rise to.

 

[cianfa72]

Which is irrelevant in this discussion because we are talking about the entire Killing vector fields, not just the vectors they evaluate to at a particular point and the integral curves those vectors give rise to.

Sorry to be pedantic, looks at the definition of subbundle for instance on J. Lee - Introduction to smooth manifolds, chapter 10.

Btw, in the 3d Euclidean space example, the set of KVFs form a linear subspace of the $\mathbb R$-vector space of smooth vector fields defined on it. It turns out that such vector subspace has dimension 6 (even though the vector space of smooth vector fields is infinite dimensional).

 

[PeterDonis]

looks at the definition of subbundle for instance on J. Lee - Introduction to smooth manifolds, chapter 10.

I don't have that reference. However, I don't think the definition of a subbundle is an issue here.

Please write down explicitly the 2D subbundle you are talking about and explain why you think it poses a problem for the claim you describe in the OP of this thread.

 

[martinbn]

Ok surely, but the point is: why such 6 spacelike KVFs (with their commutation relations) span a subbundle of dimension 3 ?

For instance on 3d Euclidean space, the 3 $\mathbb R$-linear independent rotating KVFs span a two dimensional subbundle (the integral submanifolds of are the 2-spheres).

You have to be carefull with the dimension of the bundles. The tangent bundle of $\mathbb R^3$ is 6 dimensional and the 2 dimensional distribution is a 5 dimentional subbundle.

 

[cianfa72]

The tangent bundle of $\mathbb R^3$ is 6 dimensional and the 2 dimensional distribution is a 5 dimentional subbundle.

Yes, however the point is that the definition of subbundle dimension seems to refer just to the dimension of the fiber (at least in J. Lee book and other sources).

 

[cianfa72]

Please write down explicitly the 2D subbundle you are talking about and explain why you think it poses a problem for the claim you describe in the OP of this thread.

Sorry, maybe I was unclear: actually it poses no problem at all. It was just an example to address the question why the 6 spacelike $\mathbb R$-linear independent KVFs for a spatial homogeneous and isotropic spacetime span at each point a 3 dimensional subbundle of the tangent bundle (i.e the spacetime foliation is given by (spacelike) hypersurfaces).

 

[martinbn]

Yes, however the point is that the definition of subbundle dimension seems to refer just to the dimension of the fiber (at least in J. Lee book and other sources).

Are you sure? Usually that is called the rank of the vector bundle.

 

[cianfa72]

Are you sure? Usually that is called the rank of the vector bundle.

Look for instance at the example 10.33 of J. Lee book:

a) If $M$ is a smooth manifold and $V$ is a nowhere-vanishing smooth vector field on $M$, then the set $D \subseteq TM$ whose fiber at each $p \in M$ is the linear span of $V_p$ is a smooth 1-dimensional subbundle of $TM$.

 

[PeterDonis]

the question why the 6 spacelike -linear independent KVFs for a spatial homogeneous and isotropic spacetime span at each point a 3 dimensional subbundle of the tangent bundle (i.e the spacetime foliation is given by (spacelike) hypersurfaces).

I'm not sure why tangent bundles and subbundles are even relevant to the OP question. A spacetime foliation is not a foliation of the tangent bundle. It's a foliation of the spacetime. To see how the 6 KVFs associated with homogeneity and isotropy generate a foliation by spacelike 3-surfaces, you have to look at how the integral curves of those KVFs behave in spacetime itself.

In other words, I'm not sure the first bolded statement in the quote above is even equivalent to the second. Nor do I think it matters, since I don't think trying to figure out the first is a good way to try to figure out the second in any case.

 

[martinbn]

Look for instance at the example 10.33 of J. Lee book:

That is probably an error in editing the book. If you look every where after and before that he says rank.

I also don't see how the terminology matters for you question. And I don't understand what the question really is!

 

[cianfa72]

I'm not sure why tangent bundles and subbundles are even relevant to the OP question. A spacetime foliation is not a foliation of the tangent bundle. It's a foliation of the spacetime.

No, it isn't a foliation of the tangent bundle itself. As you said it is a foliation of the spacetime as manifold.

To see how the 6 KVFs associated with homogeneity and isotropy generate a foliation by spacelike 3-surfaces, you have to look at how the integral curves of those KVFs behave in spacetime itself.

Maybe I was sloppy/unclear or you are missing the point. The subbundle enters the "picture" since it refers to a distribution over the spacetime as manifold. The 6 independent KVFs for a spatially homogeneous and isotropic spacetime define a (spacelike) smooth distribution over it (smooth since the KVFs are). Furthermore such a distribution is integrable since the Lie bracket commutators of the KVFs close.

Next the question: having 6 KVFs for a manifold of dimension 4, the dimension of the distribution they define/span can be at most 4. So, why the dimension of such (spacelike) distribution turns out to be 3 and not, for instance, 1, 2 ? That's the point.

 

[martinbn]

Next the question: having 6 KVFs for a manifold of dimension 4, the dimension of the distribution they define/span can be at most 4. So, why the dimension of such (spacelike) distribution turns out to be 3 and not, for instance, 1, 2 ? That's the point.

I answered that, because at each point they span a three dimensional subspace of the tangent space.

 

[cianfa72]

I answered that, because at each point they span a three dimensional subspace of the tangent space.

Yes, but my question is why. Is that related to the relations that exist between the Lie commutator of the 6 independent KVFs (i.e. to the structure of their Lie algebra) ?

 

[martinbn]

Yes, but my question is why. Is that related to the relations that exist between the Lie commutator of the 6 independent KVFs (i.e. to the structure of their Lie algebra) ?

He states a theorem that on a $d$ dimensional manifold you can have at most $\frac{d(d+1)}2$ symmetries. If the foliation has less than three dimensional leaves then the restriction to one of those submanifolds will give you too many (six) symmetries.

 

[cianfa72]

He states a theorem that on a $d$ dimensional manifold you can have at most $\frac{d(d+1)}2$ symmetries. If the foliation has less than three dimensional leaves then the restriction to one of those submanifolds will give you too many (six) symmetries.

Ah ok, very good.

Just to reiterate: suppose the distribution given by those 6 KVFs was only 2 dimensional (i.e. the integrable subbundle had rank 2). Then the maximum number of symmetries in any (spacelike) 2D leaf (submanifold) of the foliation would be 2*3/2=3. Therefore on each of them couldn't exist six symmetries (too many). Even more in case of 1 dimensional distribution.

 

[cianfa72]

Just to complete the argument above.

Consider the 3d euclidean space in which there are 6 $\mathbb R$-linear independent KVFs. In this case their span at each point is 3 dimensional (i.e. they span the entire tangent space at any point). The restriction to a 2-sphere let's say "supports" only 3 symmetries of them (2*3/2=3).

Instead, in the spatially homogeneous and isotropic spacetime case, by its very definition, all the 6 symmetries must apply to each leaf of the (spacelike) 3d foliation. I believe this is essentially "encoded" in the Lie algebra of these KVFs (i.e. its structure constants).

For instance the Lie algebra of the KVFs of 3d euclidean space should be different from that of the KVFs of a spatially homogeneous and isotropic spacetime.

 

[PeterDonis]

The subbundle enters the "picture" since it refers to a distribution over the spacetime as manifold.

I'm not sure what this means, but suppose it's true. So what? What does this have to do with the question you asked in the OP of this thread?

 

[cianfa72]

So what? What does this have to do with the question you asked in the OP of this thread?

Maybe I'm not able to explain my point. I used the term subbundle to mean a (smooth) distribution (as many sources imply). Full stop.

Post #19 answered my question (with some clarification/comment in my following post#20 and #21).