I Is the FLRW Metric the Only Form for One-Dimensional Maximally Symmetric Spaces?

[davidge]

This is a 1-dimensional manifold, and we've already shown that maximal symmetry is trivial for 1-dimensional manifolds.

Oh, excuse me

You're not making sense. The fact that maximal symmetry is trivial for one-dimensional manifolds does not have anything to do with dropping one dimension for ease of visualization.

Ok. So there's no subspaces of $\mathbb{R}^3$ with maximal symmetry other than those three considered in the FLRW metric?

 

[davidge]

How can we check that? Maybe trying to find the maximal number of Killing vectors for whatever space we choose and noting that that number will not be equal to three?

 

[PeterDonis]

So there's no subspaces of $\mathbb{R}^3$ with maximal symmetry other than those three considered in the FLRW metric?

The three "spaces" in the FRW metric are not subspaces of $\mathbb{R}^3$. They are just 3-dimensional spaces (one of which happens to be $\mathbb{R}^3$ with the usual flat Euclidean metric on it). And yes, they are the only 3-dimensional spaces which are maximally symmetric. Carroll proves this in Chapter 8 of his notes.

 

[davidge]

The three "spaces" in the FRW metric are not subspaces of $\mathbb{R}^3$

But can't they even be defined in terms of $\mathbb{R}^3$? (And after having defined them, we regard them as topological spaces on their own right.)

 

[PeterDonis]

How can we check that? Maybe trying to find the maximal number of Killing vectors for whatever space we choose and noting that that number will not be equal to three?

You could try that, but since there are an infinite number of possible spaces to check, it might take a while.

A better method is to find a general method that gives you all possible maximally symmetric spaces, and then just counting how many you get. That is what Carroll does in Chapter 8 of his notes.

 

[PeterDonis]

can't they even be defined in terms of $\mathbb{R}^3$?

No. For someone who said they had concerns about using the notion of embedding, you seem to have a big problem getting rid of the conceptual crutch of trying to visualize everything in terms of embeddings/subspaces/whatever using Euclidean spaces. You need to stop doing that, and learn how to deal with these various spaces as metric spaces in their own right, using only their intrinsic properties.

To restate the one part of this that does have some relationship to $\mathbb{R}^3$: of the three possible cases, the $k = 0$ case and the $k = -1$ case can be completely covered by a single coordinate chart (for the $k = 0$ case this is trivial since it is just $\mathbb{R}^3$; for the $k = -1$ case constructions are easy to find), which means there is a one-to-one mapping between those spaces and $\mathbb{R}^3$. For the $k = 1$ case, however, this is not possible--which should also be obvious because $\mathbb{S}^3$ is compact and $\mathbb{R}^3$ is not, so there can't possibly be a one-to-one mapping between them. It takes at least two coordinate charts to cover $\mathbb{S}^3$.

 

[PeterDonis]

can't they even be defined in terms of $\mathbb{R}^3$?

Note also that, even if we allow ourselves to use embeddings, you have to use a Euclidean space of more than 3 dimensions to construct an embedding for hyperbolic 3-space or the 3-sphere.

 

[davidge]

You could try that, but since there are an infinite number of possible spaces to check, it might take a while.

For someone who said they had concerns about using the notion of embedding, you seem to have a big problem getting rid of the conceptual crutch of trying to visualize everything in terms of embeddings/subspaces/whatever using Euclidean spaces. You need to stop doing that, and learn how to deal with these various spaces as metric spaces in their own right, using only their intrinsic properties.

Can you show me the definition of, say, $\mathbb{S}^3$ without talking about $\mathbb{R}^4$?

 

[PeterDonis]

Can you show me the definition of, say, $\mathbb{S}^3$ without talking about $\mathbb{R}^4$?

I can, but I won't. You need to learn the basic topology for yourself. Then you won't make so many mistaken assumptions. Any basic topology textbook will answer this question for you.

(Note that many textbooks which are not focused on topology but make use of it--which includes many, if not most, physics textbooks on GR--will give you definitions that do make use of an embedding. You need to look at a rigorous topology textbook that does it correctly. Often, to make sure you are getting a definition that makes the minimum possible assumptions, you need to look to mathematicians, not physicists; physicists--which often includes me--are notoriously sloppy about these things. )

 

[davidge]

So can you point to me some of such books, so that I can look for them...

you need to look to mathematicians, not physicists; physicists--which often includes me--are notoriously sloppy about these things

Never the less I believe such concepts as the one we have been discussing here are extremely important and should be in the mind of any physicist who claim knowing the fundamentals of physics.

 

[PeterDonis]

can you point to me some of such books, so that I can look for them...

I don't have any particular references handy, unfortunately. For a quick overview of the kinds of topological constructions I'm referring to, see here:

https://en.wikipedia.org/wiki/3-sphere#Topological_construction

 

[davidge]

see here: https://en.wikipedia.org/wiki/3-sphere#Topological_construction

I'm sorry, but I can't see in that article any definition of $\mathbb{S}^3$ which does not depend on $\mathbb{R}^4$. The very definition they give uses $\mathbb{R}^4$: https://en.wikipedia.org/wiki/3-sphere#Definition

 

[davidge]

I think we are missing the main point of the thread. So let me see if I got things correctly:

I asked in the OP whether or not that given form of the metric is the only possible form. You said in later post that it's the unique possible form, because the hyperbolic space, the spherical space and the plane space are the ones that can have the maximal number of Killing vectors, and that it's explained in Carroll's notes how to prove that. Correct?

 

[PeterDonis]

I'm sorry, but I can't see in that article any definition of $\mathbb{S}^3$ which does not depend on $\mathbb{R}^4$.

The two constructions in the specific part of the article I linked you to--the gluing construction and the one-point compactification--both do not depend on $\mathbb{R}^4$.

 

[PeterDonis]

I asked in the OP whether or not that given form of the metric is the only possible form. You said in later post that it's the unique possible form

No, that's not what I said. You actually asked two different questions in the OP, and I only answered the first one.

First, you asked if Euclidean 3-space, hyperbolic 3-space, and the 3-sphere are the only possible maximally symmetric 3-dimensional spaces. I answered yes to that, and referred you to Carroll's proof.

Second, you asked if the form of the metric you wrote down in the OP is the only possible form that can describe those spaces. (I am assuming that we are actually talking about the correct 3-dimensional metric, not the one-dimensional version you wrote in the OP.) You appear to be assuming that that's the same question as the first one above, but it isn't. There are multiple possible coordinate charts you can choose to represent these spaces, which lead to different expressions for the metric; the form of the metric you wrote in the OP is only one of them. So the answer to the second question is no.

 

[PeterDonis]

I believe such concepts as the one we have been discussing here are extremely important and should be in the mind of any physicist who claim knowing the fundamentals of physics.

They are. But physicists writing for other physicists, or writing textbooks to teach physics, know that the most useful versions of the concepts they use aren't always the same as the ones that are most easily derived with full mathematical rigor. That's why, for example, physicists often use embeddings of particular spaces in higher dimensional spaces--because it makes it easier to work with those spaces in a practical sense. But the physicists know perfectly well that, logically speaking, the existence of the spaces does not depend on the existence of the embeddings; that's why they don't worry about making practical use of the embeddings: because it doesn't commit them to saying that the higher dimensional spaces used in the embeddings must be "real". It's just an aid to practical calculation or visualization.

 

[davidge]

The two constructions in the specific part of the article I linked you to--the gluing construction and the one-point compactification--both do not depend on $\mathbb{R}^4$.

I think I should clarify what I mean a bit more. What I mean is that there seems no way of defining a metric for $\mathbb{S}^3$ without talking about $\mathbb{R}^4$, that is, without talking about an embedding. Is there any way?

No, that's not what I said. You actually asked two different questions in the OP, and I only answered the first one.

oh that's correct, excuse me

 

[PeterDonis]

Is there any way?

Of course: just construct the topological space $\mathbb{S}^3$, using either of the two constructions I linked to (gluing or one-point compactification), and then find the unique metric on that topological space that gives it a constant curvature.

 

[davidge]

topological space that gives it a constant curvature

So the space must have constant curvature?

 

[PeterDonis]

So the space must have constant curvature?

You said you were talking about the 3-sphere $\mathbb{S}^3$ as a metric space, which is defined as the metric space with topology $\mathbb{S}^3$ and constant curvature.

If you are asking, does a maximally symmetric space have to have constant curvature, see if you can deduce the answer from the definition of maximally symmetric space. (Hint: Carroll's discussion in Chapter 8 of his notes is helpful here.)

 

[davidge]

If you are asking, does a maximally symmetric space have to have constant curvature, see if you can deduce the answer from the definition of maximally symmetric space. (Hint: Carroll's discussion in Chapter 8 of his notes is helpful here.)

I think the answer is "yes", not deduced from Carroll's notes, but I remember seen this discussion on Weinberg's book.

If I remember well, constant curvature is a consequence of a maximally symmetric space.

 

[davidge]

constant curvature

In the case in question (FLRW metric), this would be so for any value of $k$. But then would this imply that even a non homogeneous space (case $k=0$) has maximal symmetry?

 

[PeterDonis]

I think the answer is "yes"

I agree.

In the case in question (FLRW metric), this would be so for any value of $k$

Yes.

would this imply that even a non homogeneous space (case $k=0$) has maximal symmetry?

The $k = 0$ case is just Euclidean 3-space; it should be obvious that this space is maximally symmetric. I don't know why you think this case is non-homogeneous.