I Galaxy recession and Universe expansion

[cianfa72]

Of SO(3) KVFs, yes. Note that the isometries associated with homogeneity, the spatial translations in a spacelike hypersurface of constant time, are also generated by a three-parameter group of KVFs, but the group is not SO(3).

Sorry, such KVFs form a Lie algebra (w.r.t. the Lie bracket). Are themselves elements of SO(3) group, or are the isometries they generate actually elements of groups each isomorphic to SO(3) ?

 

[PeterDonis]

such KVFs form a Lie algebra

Sure, any group of KVFs forms a Lie algebra.

Are themselves elements of SO(3) group

A Lie algebra is not the same as a Lie group. Elements of a Lie algebra can be generators of a Lie group.

or are the isometries they generate actually elements of groups each isomorphic to SO(3) ?

It would be really helpful if you would be specific about which isometries you are talking about.

In FRW spacetime, any spacelike surface of constant FRW coordinate time $t$ has two three-parameter groups of isometries, generated by two three-parameter groups of KVFs.

One three-parameter group, the one associated with isotropy at every point, is associated with the group SO(3).

The other three-parameter group, the one associated with homogeneity, is associated with a group that depends on the specific FRW model. In a spatially flat model, the group is E(3) (I think that's right--it's the three-parameter group of translations in Euclidean 3-space). In a spatially open model, the group is H(3) (again, I think that's right--it's the three-parameter group of translations in hyperbolic 3-space). In a spatially closed model, I believe the group is SO(4).

 

[cianfa72]

A Lie algebra is not the same as a Lie group. Elements of a Lie algebra can be generators of a Lie group.

Yes, so the "set" of three-parameters KVFs associated with the isotropy at every point (on any spacelike hypersurface of constant FRW coordinate time) form a Lie algebra and generate isometries that form a Lie group isomorphic to SO(3).

One three-parameter group, the one associated with isotropy at every point, is associated with the group SO(3).

Yes, I was referring to this three-parameter group.

 

[cianfa72]

Sorry to bother you with these details. We said FRW spacetime is spherically symmetric regardless the point we choose as center of symmetry. The geometry of any spacelike hypersurface of constant cosmological time can be only one of the three type: flat (Euclidean), hyperbolic, spherical.

In any case there will be 6 linearly independent KVFs (one three-parameters "set" of KVFs for translations and one three--parameter "set" of KVFs for rotations about one chosen point).

The notion of linearly independent KVFs actually means that all the coefficients of their linear combination that give the null vector field are null (zero).

Now each other "set" of KVFs that are generators of the isometry group of rotations about any other point on that manifold can be given as linear combination of the first 6 KVFs. In other words, from the three-parameters translational KVFs and from the "set" of three-parameters rotational KVFs about one point, we get the set of three-parameters rotational KVFs about any other point.

 

[PeterDonis]

We said FRW spacetime is spherically symmetric regardless the point we choose as center of symmetry.

Yes.

The geometry of any spacelike hypersurface of constant cosmological time can be only one of the three type: flat (Euclidean), hyperbolic, spherical.

Yes.

In any case there will be 6 linearly independent KVFs (one three-parameters "set" of KVFs for translations and one three--parameter "set" of KVFs for rotations about one chosen point).

Yes.

The notion of linearly independent KVFs actually means that all the coefficients of their linear combination that give the null vector field are null (zero).

I'm not sure what you mean by this. "Linearly independent" means that none of the KVFs can be expressed as a linear combination of the others.

Now each other "set" of KVFs that are generators of the isometry group of rotations about any other point on that manifold can be given as linear combination of the first 6 KVFs. In other words, from the three-parameters translational KVFs and from the "set" of three-parameters rotational KVFs about one point, we get the set of three-parameters rotational KVFs about any other point.

No, this is not correct. A linear combination of a translation and a rotation is not a rotation about a different point.

What is correct is that you can Fermi-Walker transport the "rotation" KVFs at one point along the integral curves of a translation to another point, and you will get the "rotation" KVFs at the other point. But that's not the same as forming a new KVF from a linear combination of the translation and rotation.

 

[cianfa72]

No, this is not correct. A linear combination of a translation and a rotation is not a rotation about a different point.

I think I misinterpreted it. The set of KVFs on a manifold is not a vector space, therefore we can't get for instance a KVF for rotation about a different point as linear combination of overall 6 translation KVFs and rotation KVFs at another point.

What is correct is that you can Fermi-Walker transport the "rotation" KVFs at one point along the integral curves of a translation to another point, and you will get the "rotation" KVFs at the other point.

Maybe the result is the same as Lie dragging the rotation KVFs at one point along the integral curves of translation to another point.

 

[PeterDonis]

The set of KVFs on a manifold is not a vector space

Correct.

 

[PeterDonis]

Maybe the result is the same as Lie dragging the rotation KVFs at one point along the integral curves of translation to another point.

Actually, Lie dragging might be the correct transport law here. I believe we had a previous thread where we showed that Lie transport and Fermi-Walker transport are not necessarily the same.

 

[cianfa72]

The set of KVFs on a manifold is not a vector space, therefore we can't get for instance a KVF for rotation about a different point as linear combination of overall 6 translation KVFs and rotation KVFs at another point.

I've some doubts about this. From the above 7 KVFs are not linearly independent (see also Sean Carroll Spacetime and geometry - section 3.9).
What does it mean ? We said that the set of KVFs on a manifold is not a vector space (w.r.t. the pointwise sum of vector fields).

Perhaps the point is that if we take 7 KVFs then we can always write down any of them by "some kind of operations" acting on the other 6 KVFs. For example, as in post #58, we get the rotation KFV about a point by Lie dragging the rotation KVFs about a different point.

On the other hand, we can always find 6 KFVs such that none of them can be obtained as linear combination of the other 5 and does not exist a "some kind of operation" to get it from the others.

 

[cianfa72]

Another point related to the definition of isotropy. As in Sean Carroll Spacetime and geometry section 8.2, an universe spatially homogeneous and isotropic can be foliated into maximally symmetric spacelike hypersurfaces .

In section 8.1 the isotropy condition at a point on any such 3D spacelike manifold $M$ goes as follows:

More formally, a manifold $M$ is isotropic around a point $p$ if, for any two vectors $V$ and $W$ in $T_pM$ there is an isometry of $M$ such that the pushforward of $W$ under the isometry is parallel with $V$ (not pushed forward).

This definition should be equivalent to the existence of spacelike KVFs associated to those isometries (the pushforward is actually the differential of the isometry map evaluated at the point $p$).

 

[PeterDonis]

7 KVFs are not linearly independent

More precisely, you can't find 7 linearly independent KVFs at a particular point in one spacelike hypersurface of constant FRW coordinate time. You can only find 6.

What does it mean ? We said that the set of KVFs on a manifold is not a vector space

A vector space is defined as a space satisfying a particular set of axioms. Look up those axioms and then check whether the set of KVFs on a manifold satisfies them.

Perhaps the point is that if we take 7 KVFs then we can always write down any of them by "some kind of operations" acting on the other 6 KVFs

If you are talking about linearly independent, look up the definition of that. That definition is not in terms of "some kind of operations". There is one particular operation involved. Look it up and see.

we get the rotation KFV about a point by Lie dragging the rotation KVFs about a different point.

This has nothing to do with either linear independence or forming a vector space. (Which are two different things, btw.)

This definition should be equivalent to the existence of spacelike KVFs associated to those isometries

Yes.

 

[cianfa72]

More precisely, you can't find 7 linearly independent KVFs at a particular point in one spacelike hypersurface of constant FRW coordinate time. You can only find 6.

Ah ok, so the the notion of linearly independence applies to KVFs "evaluated" at specific points.

A vector space is defined as a space satisfying a particular set of axioms. Look up those axioms and then check whether the set of KVFs on a manifold satisfies them.

Re-thinking about it, I believe the set of KVFs on a manifold is a real vector space. Namely the pointwise sum of two KVFs is a KFV (i.e. the sum is closed) as the scalar moltiplication for a real number (together with the Lie bracket such vector space is promoted to an algebra).

 

[PeterDonis]

the notion of linearly independence applies to KVFs "evaluated" at specific points.

No, it applies to vector fields in general. The reason I said "at a specific point" is that the KVFs associated with isotropy at different points are different KVFs.

Re-thinking about it, I believe the set of KVFs on a manifold is a real vector space. Namely the pointwise sum of two KVFs is a KFV (i.e. the sum is closed) as the scalar moltiplication for a real number (together with the Lie bracket such vector space is promoted to an algebra).

Those aren't the only vector space axioms.

 

[cianfa72]

Those aren't the only vector space axioms.

I could be wrong but an algebra requires an underlying structure of vector space on a field -- Lie algebra.

 

[PeterDonis]

I could be wrong but an algebra requires an underlying structure of vector space on a field -- Lie algebra.

That is true, but irrelevant to what I said.

 

[cianfa72]

That is true, but irrelevant to what I said.

Sorry, I don't understand. If the set of KVFs on a manifold is a Lie algebra then it is a vector space on $\mathbb R$ w.r.t. the pointwise sum of KVFs.

 

[PeterDonis]

If the set of KVFs on a manifold is a Lie algebra then it is a vector space

Ah, I see. Yes, that's correct.

However, I still think it would be a good idea for you to check that the set of KVFs satisfies all of the vector space axioms, not just the ones you mentioned.

 

[cianfa72]

However, I still think it would be a good idea for you to check that the set of KVFs satisfies all of the vector space axioms, not just the ones you mentioned.

My point is the following: start by noting that vector fields on a manifold form themselves a vector space on $\mathbb R$, therefore it suffices to show that KVFs are closed both w.r.t. the vector field sum and the product for a scalar in $\mathbb R$.

We can use the following definition: $X$ is a KVF iff:
$$g(\nabla_YX,Z)+g(Y,\nabla_ZX)=0,\ \forall Y,Z\in\Gamma(TM).$$ Using the linear property of covariant derivatives and bi-linearity of metric tensor $g$ we can show that both the aforementioned conditions are satisfied.

 

[PeterDonis]

vector fields on a manifold form themselves a vector space on $\mathbb R$, therefore it suffices to show that KVFs are closed both w.r.t. the vector field sum and the product for a scalar in $\mathbb R$.

No, it doesn't. You also have to show that the identity element of the vector space is in the set of KVFs. That might seem trivial, but it at the very least should give some food for thought.

 

[cianfa72]

No, it doesn't. You also have to show that the identity element of the vector space is in the set of KVFs. That might seem trivial, but it at the very least should give some food for thought.

Yes, the identity element of the vector field vector space is the null vector field $0$. If we "insert" it into post#68 definition we get $$\nabla_Y0 =0, \nabla_Z0=0,\ \forall Y,Z\in\Gamma(TM)$$ hence $$g(0,Z)=0, g(Y,0), \ \forall Y,Z \in\Gamma(TM)$$

 

[PeterDonis]

Yes, the identity element of the vector field vector space is the null vector field $0$. If we "insert" it into post#68 definition we get $$\nabla_Y0 =0, \nabla_Z0=0,\ \forall Y,Z\in\Gamma(TM)$$ hence $$g(0,Z)=0, g(Y,0), \ \forall Y,Z \in\Gamma(TM)$$

Yes. What is the physical meaning of the null vector field, considered as a Killing vector field?

 

[cianfa72]

What is the physical meaning of the null vector field, considered as a Killing vector field?

I don't know exactly Maybe is this: if we do not move in any direction from a point the metric can't change.

 

[PeterDonis]

if we do not move in any direction from a point the metric can't change.

Yes.

 

[cianfa72]

Coming back to post#61 now the situation is: on each 3D spacelike hypersurface maximally symmetric there are 6 linearly independent KVFs and 7 KVFs are always linearly dependent. That means there exists a basis of 6 KVFs.

I think the above result implies that a rotation KVFs about a given point on the spacelike hypersurface can always be written as linear combination of 6 linearly independent KVFs.

 

[PeterDonis]

on each 3D spacelike hypersurface maximally symmetric there are 6 linearly independent KVFs and 7 KVFs are always linearly dependent.

Yes.

That means there exists a basis of 6 KVFs.

Yes, but you can still split them up further into subspaces. See below.

I think the above result implies that a rotation KVFs about a given point on the spacelike hypersurface can always be written as linear combination of 6 linearly independent KVFs.

In the absence of additional structure, that would be the best you could do, yes. But there is additional structure. The rotations centered on a given point form their own closed subspace. So do the translations. So if we pick three linearly independent rotation KVFs, they form a basis by themselves of the subspace of rotations. Similarly, if we pick three linearly independent translation KVFs, they form a basis by themselves of the subspace of translations. Only KVFs that are neither pure rotations nor pure translations would need the full basis of 6 linearly independent KVFs to express them.

 

[cianfa72]

The rotations centered on a given point form their own closed subspace. So do the translations. So if we pick three linearly independent rotation KVFs, they form a basis by themselves of the subspace of rotations. Similarly, if we pick three linearly independent translation KVFs, they form a basis by themselves of the subspace of translations. Only KVFs that are neither pure rotations nor pure translations would need the full basis of 6 linearly independent KVFs to express them.

Ok so, fixed a point on the spacelike manifold, the linear subspace of rotation KVFs around it is spanned by 3 linearly independent rotation KVFs around it.

Furthermore the subspace of rotation KVFs around a point can be always given as linear combination of any 3 linearly independent rotation KVFs (e.g. as the linear combination of 3 linearly independent KVFs around a different point).

 

[PeterDonis]

fixed a point on the spacelike manifold, the linear subspace of rotation KVFs around it is spanned by 3 linearly independent rotation KVFs around it.

Yes.

the subspace of rotation KVFs around a point can be always given as linear combination of any 3 linearly independent rotation KVFs (e.g. as the linear combination of 3 linearly independent KVFs around a different point).

No.

 

[cianfa72]

No.

This is the point unclear to me. In post#75 you said

So if we pick three linearly independent rotation KVFs, they form a basis by themselves of the subspace of rotations

Three linearly independent rotation KVFs around a point are not 3 linearly independent rotation KVFs for the manifold ?

 

[PeterDonis]

Three linearly independent rotation KVFs around a point are not 3 linearly independent rotation KVFs for the manifold ?

What do you mean by "for the manifold"? The set of rotations about a given point induces a set of isometries on the entire manifold. It's just a different set of isometries from the set induced by the set of rotations about a different point.

 

[cianfa72]

The set of rotations about a given point induces a set of isometries on the entire manifold. It's just a different set of isometries from the set induced by the set of rotations about a different point.

Yes, but if the set of KVFs on a maximally symmetric 3D spacelike manifold is a vector space of dimension 6 then the rotation KVFs about a given point must be writable as linear combination of any 6 linear independent KVFs.

Are the rotation KVFs around a point in the same KVFs subspace as the rotation KVFs around a different point ?

 

[PeterDonis]

if the set of KVFs on a maximally symmetric 3D spacelike manifold is a vector space of dimension 6 then the rotation KVFs about a given point must be writable as linear combination of any 6 linear independent KVFs.

Yes, but not all of them will be rotation KVFs if you pick a basis that includes rotation KVFs about a different point.

 

[cianfa72]

Yes, but not all of them will be rotation KVFs if you pick a basis that includes rotation KVFs about a different point.

You mean not all the KVFs in the picked basis (that includes rotation KVFs about a different point) will be rotation KVFs from the point of view of the given point.

 

[PeterDonis]

You mean not all the KVFs in the picked basis (that includes rotation KVFs about a different point) will be rotation KVFs from the point of view of the given point.

None of the KVFs in the picked basis will be pure rotation KVFs about the given point if the basis is picked using rotation KVFs centered on a different point.

Note that three of the KVFs in any basis will not be pure rotation KVFs; in the simplest case they will be pure translation KVFs.

 

[cianfa72]

Note that three of the KVFs in any basis will not be pure rotation KVFs; in the simplest case they will be pure translation KVFs.

Correct me whether I am wrong. A pure rotation KFV about a point evaluated at that point is the null vector (the 3 pure translation KVFs evaluated at that point form a basis for the tangent space there).

Btw what I quoted from you applies to any other basis of KVFs different from the basis in which the 3 KVFs "act" as pure rotation KVFs about the given point.

 

[PeterDonis]

A pure rotation KFV about a point evaluated at that point is the null vector.

"Centered" on a particular point is not the same thing as evaluated at that point. You need to take a step back and think more carefully.

Take a simpler example: the 2-dimensional Euclidean plane. This manifold has a three-parameter group of KVFs, or at least that's the way it is usually stated. This can be broken up into a two-parameter group of translations and, as it's usually stated, one rotation, corresponding to an SO(2) isometry on the manifold. But let's unpack that further.

Choose standard Cartesian coordinates on the plane, which requires picking one particular point as the origin. That point then becomes the "center" of the SO(2) rotation isometry.

Now write down a basis for the three-parameter group of KVFs consisting of the translations and the rotations centered on the chosen origin. It looks like this:

$$
\partial_x
$$
$$
\partial_y
$$
$$
- y \partial_x + x \partial_y
$$

It should be easy for you to verify that all three of these are KVFs and that they are linearly independent.

Now, looking at the third KVF above, the "rotation" KVF, you can see that it vanishes at the origin, $(x, y) = (0, 0)$. In other words, if you evaluate it at the center point of rotation, it vanishes. Or, to put it the way you correctly put it in a previous post, it doesn't "go anywhere" at the center point--it maps the center point to itself. But it doesn't vanish at any other point, and it obviously induces an isometry on the entire manifold that does not map every point on the manifold to itself.

Now suppose we decide to look at a rotation about a different center point, say the point $(x, y) = (1, 1)$. The KVF that corresponds to this rotation, in the coordinates we used above, would be:

$$
- (y - 1) \partial_x + (x - 1) \partial_y
$$

This is, as should be evident, a linear combination of all three of the KVFs given above. In other words, it is not a pure rotation about the center point we chose above, the point $(x, y) = (0, 0)$. It's a linear combination of a rotation about that point, and two translations (in the $x$ and $y$ directions).

 

[cianfa72]

Thanks @PeterDonis it makes sense. From the point of view of isotropy at a point $p$, as discussed from Sean Carroll in his book - section 8.1, given two vectors $V$ and $W$ in the tangent space at $p$ must exist a one-parameter diffeomorphism $\phi_t$ that is an isometry such that its pushforward $\phi_*$ (i.e. the differential map $d\phi$) evaluated at $p$ maps $V$ into a vector parallel to $W$. This one-parameter diffeomorphism (isometry) maps $p$ into itself and defines a rotation KFV about the point $p$.

 

[cianfa72]

No, this is not correct. A linear combination of a translation and a rotation is not a rotation about a different point.

Coming back to this post, your point was that not any linear combination of translation KFVs and rotation KVFs about a point results in a rotation KVF about a different point. However there is a specific linear combination such that the above is true.

 

[PeterDonis]

Coming back to this post, your point was that not any linear combination of translation KFVs and rotation KVFs about a point results in a rotation KVF about a different point. However there is a specific linear combination such that the above is true.

Yes.

 

[cianfa72]

Ok, anyway as you said in post#58, for a maximally symmetric manifold we can get a rotation KVFs about a point by Lie dragging along the translation KVFs the rotation KVFs about a different point.

 

[PeterDonis]

for a maximally symmetric manifold we can get a rotation KVFs about a point by Lie dragging along the translation KVFs the rotation KVFs about a different point.

Intuitively it seems like that should be the case. However, I have not done an actual computation to confirm it. You might try checking it for the case of 2-D Euclidean space; see if Lie dragging the rotation KVF centered on $(0, 0)$ along an appropriate translation KVF gives the rotation KVF centered on $(1, 1)$.

 

[PeterDonis]

Moderator's note: A subthread based on an unacceptable reference has been deleted. Please do not give pop science articles as references.